#$&* course giancoli text goals in c:\labrynth of netbook and e: (hopefully to be copied to shell_01 and thinkpad)
.............................................
Given Solution: `a** log{base 2}(x) = 4 translates to 2^4 = x The value of x is therefore 2^4 = 16. NEARLY-CORRECT STUDENT SOLUTION log{base 2}(x) = 4 translates to 2^x = 4, and 2^2 = 4 so x = 2. INSTRUCTOR RESPONSE You have the right idea, but you confused the role of x when you translated log{base 2}(x) = 4. As you will see below, the correct translation would be 2^4 = x. It's very easy to get the variables reversed, so the translation has to be done carefully. It's best if you understand the reasons behind the translation, so the following explanations include the reasoning. One solution: The y = log{base b} function is inverse to the y = b^x function. This means that if you reverse the columns of the y = b^x table, you get the y = log{base b}(x) table. When you reverse columns, you are reversing the x and y variables. For this reason, y = log{base b}(x) means the same thing as b^y = x. Whether you completely understand about reversing the columns, etc., be sure you remember this. You are asked to find the x value for which y = log{base 2}(x). The value of y = log{base 2}(x) is y. So when the value is 4, this means that 4 = log{base 2}(x). Now, since y = log{base b}(x) means the same thing as b^y = x , the equation 4 = log{base 2}(x) means the same thing as 2^4 = x. Since 2^4 = 16, we conclude that x = 16. Alternative solution: We can make a table for y = log{base 2}(x) y = log{base 2}(x) is defined at the inverse of the function of y = 2^x. We start with a table for y = 2^x: x y = 2^x -2 1/4 -1 1/2 0 1 1 2 2 4 3 8 4 16 5 32 We reverse the columns of the table, obtaining a table for y = log{base 2}(x) x y = log{base 2}(x) 1/4 -2 1/2 -1 1 0 2 1 4 2 8 3 16 4 32 5 You should sketch rough graphs of these two functions. From the tables, and from the graphs, it should be clear that y = log{base 2}(x) first takes value 4 when x = 16. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qfor what value of x will the function y = ln(x) first reach y = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 = ln(x) e^4 = x x = 54.598 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `ay = ln(x) means that e^y = x. The function y = ln(x) will first reach y = 4 when x = e^4 = 54.6 approx. ** ln(x) = 4 translates to x = e^4, which occurs at x = 55 approx. ln(x) = 2 translates to x = e^2, which occurs at x = 7.4 approx. ln(x) = 3 translates to x = e^3, which occurs at x = 20 approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q3. Explain why the negative y axis is an asymptote for a log{base b}(x) function explain why this is so only if b > 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = b^x is inverse. The smaller the x value in the negative range, the closer the graph is to the negative x axis. If x and y columns are reversed to fit the log{base b}(x) then the graph has a asymptote to the neg. y axis. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** the log{base b} function is the inverse of the y = b^x function. Assuming b > 1, large negative values of x lead to positive b^x values near zero, resulting in a horizontal asymptotes along the negative x axis. When the columns are reversed we end up with small positive values of x associated with large negative y, resulting in a vertical asymptote along the negative y axis. You can take a negative power of any positive b, greater than 1 or not. For b > 1, larger negative powers make the result smaller, and the negative x axis is an asymptote for the function y = b^x. For b < 1, larger positive powers make the result smaller, and the positive x axis becomes the asymptote. ** STUDENT COMMENT I understand it to read it but cant remember it after I have read it. INSTRUCTOR RESPONSE Let's consider some of the statements in the given solution. First statement: Assuming b > 1, large negative values of x lead to positive b^x values near zero, resulting in a horizontal asymptotes along the negative x axis. Understanding what that means: Pick a value for b. This statement assumes b > 1, so let's pick something simple like b = 2. b = 3, or b = 10, or whatever, would also work, but we'll just use b = 2. The statement says 'large negative values of x lead to positive b^x values near zero'. So let's test this for b = 2. What happens to the value of 2^x for increasing negative values of x? For x = -1, -2, -3, -4 the values of b^x are 2^-1 = 1/2, 2^-2 = 1/4, 2^-3 = 1/8, 2^-4 = 1/16. Now 1/16 is clearly positive, and clearly pretty close to zero. And we aren't even into very large negative values of x. If we continue to x = -5, -6, -7, ..., our y = 2^x values stay positive, but keep getting closer and closer to 0. By the time x = -10, our y value is 1/1024, positive but less that .001. There's no limit to how close our values can get to zero. However they always stay positive. Now if you plot y vs. x for x = -1, -2, -3 and -4, you will see how the asymptote forms. As you move to the left along the negative x axis, the y values keep getting smaller and smaller. The graph continues to approach, but never reaches, the negative x axis. A table of values for y = 2^x: x y = 2^x -4 1/16 -3 1/8 -2 1/4 -1 1/2 0 1 1 2 We chose b = 2 as a basis for this discussion. You might want to see what happens if b = 3, and if b = 10. You will find that the y values again stay positive but as we move through x values -1, -2, -3, -4, ..., we approach zero much more quickly than for b = 2. We still have an asymptote at the negative x axis, but the graph approaches it much more quickly for larger values of b. Second statement: When the columns are reversed we end up with small positive values of x associated with large negative y, resulting in a vertical asymptote along the negative y axis. Understanding what this means: We are talking about reversing the columns of our y = b^x table. For our previous b = 2 example, the reversed table is the function y = log{base b}(x), for b = 2: x y = log{base 2}(x) 1/16 -4 1/8 -3 1/4 -2 1/2 -1 1 0 2 1 Graph this. You will see how the asymptote forms with the negative y axis. Statement 3: For b < 1, larger positive powers make the result smaller, and the positive x axis becomes the asymptote. Understanding what this means: Pick a positive value of b which is less than 1. Here we'll pick b = 1/2. What happens to the function y = (1/2)^x for larger and larger positive values of x? For x = 1, 2, 3, 4, ..., we find that y = (1/2)^x takes values 1/2, 1/4, 1/8, 1/16, ... . These are the same values obtained in the preceding example y = 2^x for x = -1, -2, -3, -4, ... . If you make a table and a graph you will see that the y = (1/2)^x function approaches an asymptote with the positive x axis. Additional insights: The same thing will happen for any value of b which is less than 1. If b is less than 1, every multiplication by b gives a smaller result than before, so the values of y = b^x continue to decrease as x increases. If you reverse the columns of the y = (1/2)^x table, you get the y = log{base 1/2)(x) table. If you graph this table, you will see how the y = (1/2)^x asymptote with the positive x axis becomes an asymptote of the log function with the positive y axis. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q5. What are your estimates for the values of b for the two exponential functions on the given graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Graphs intersect at point (0,1). The blue graphs other y value 1 unit away is approx. 3.2, so its point is (1, 3.3). the pink graph 1 unit away point is approx. (1, 7.2) both 1 unit from their basic points. y coordinates of second sets are A*b so the b values are 3.3/1=3.2 and 7.2/1 = 7.2 respectfully. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The exponential function y = A b^x has y intercept (0, A) and basic point (1, A * b). Both graphs pass thru (0, 1) so A = 1. The x = 1 points are (1, 3.5) and (1, 7.3) approx.. Thus for the first A * b = 3.5; since A = 1 we have b = 3.5. For the second we similarly conclude that b = 7.3. So the functions are y = 3.5^x and y = 7.3^x, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### Differences are because of human error in viewing the graphs#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qAt what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using solutions given in previous answer, y = 3.5^x: (2, .553), (3, .876), (4, 1.108) y = 7.3^x: (2, .348), (3, .552), (4, .697) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Using trial and error we find that for y = 3.5^x, the x values .55, .88 and 1.11 give us y = 2, 3 and 4. We find that for y = 7.3^x, the x values .35, .55 and .70 give us y = 2, 3 and 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(10,000) = 4 dB = 10 * 4 = 40 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `adB = 10 log(I / I0). If the sound is 10,000 times as loud as hearing threshold intensity then I / I0 = 10,000. log(10,000) = 4, since 10^4 = 10,000. So dB = 10 log(I / I0) = 10 * 4 = 40. STUDENT QUESTION I didnt understand were log(10000) is 4 came from i think i would understand the problem if i new how to come up with that INSTRUCTOR RESPONSE The reason is in the line 'log(10,000) = 4, since 10^4 = 10,000' Recall that log(x) and 10^x are inverse functions. So 10^4 = 10 000 means the same thing as log(10 000) = 4. The formal definition is in terms of inverse functions. You can also think of the log of a number as the power to which 10 must be raised to get the number. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(100) = 2, 10*2=20dB log(10,000,000) = 7, 10*7=70dB log(1,000,000,000) = 9, 10 * 9 = 90dB confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a10 log(100) = 10 * 2 = 20, so a sound which is 100 times hearing threshold intensity is a 20 dB sound. 10 log(10,000,000) = 10 * 7 = 70, so a sound which is 100 times hearing threshold intensity is a 70 dB sound. 10 log(1,000,000,000) = 10 * 9 = 90, so a sound which is 100 times hearing threshold intensity is a 90 dB sound. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qhow can you easily find these decibel levels without using a calculator? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In the previous problem the log solutions corresponds to the number of zeros on the numbers multiples of 10. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSince 10^1, 10^2, 10^3, ... are 10, 100, 1000, ..., the power of 10 is the number of zeros in the result. Since the log of a number is the power to which 10 must be raised to get this number, the log of one of these numbers is equal to the number of zeros. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10 log(500) = 26.99 dB 10 log(30,000,000) = 74.77 dB 10 log(7,000,000,000) = 98.45 dB confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a10 log(500) = 10 * 2.699 = 26.99 so this is a 26.99 dB sound. 10 log(30,000,000) = 10 * 7.477 = 74.77 so this is a 74.77 dB sound. 10 log(7,000,000,000) = 10 * 9.845 = 98.45 so this is a 98.45 dB sound. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 40 = 10 log(I/Io) 4 = log(I/Io) I/Io = 1000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Let x be the ratio I / I0. Then we solve the equation 40= 10*log(x). You solve this by dividing by 10 to get log(x) = 4 then translating this to exponential form x = 10^4 = 10,000. The sound is 10,000 times the hearing threshold intensity, so I = 10,000 I0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####I should have multiplied both sides by the Io for the last step.#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qAnswer the same question for sounds measuring 20, 50, 80 and 100 decibels. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 20 = 10 log(I/Io) 2 = log(I/Io) I = 100 Io 50 = 10 log(I/Io) 5 = log(I/Io) I = 100,000 Io 80 = 10 log(I/Io) 8 = log(I/Io) I = 100,000,000 Io 100 = 10 log(I/Io) 10 = log(I/Io) I = 10,000,000,000 Io confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** since dB = 10 log(I / I0) we have log(I/I0) = dB / 10. Translating to exponential form this tells us that I / I0 = 10^(dB/10) wo that I = I0 * 10^(dB/10). For a 20 dB sound this gives us I = I0 * 10^(20/10) = I0 * 10^2 = 100 I0. The sound is 100 times the intensity of the hearing threshold sound. For a 50 dB sound this gives us I = I0 * 10^(50/10) = I0 * 10^5 = 100,000 I0. The sound is 100,000 times the intensity of the hearing threshold sound. For an 80 dB sound this gives us }I = I0 * 10^(80/10) = I0 * 10^8 = 100,000,000 I0. The sound is 100,000,000 times the intensity of the hearing threshold sound. For a 100 dB sound this gives us I = I0 * 10^(100/10) = I0 * 10^10 = 10,000,000,000 I0. The sound is 10,000,000,000 times the intensity of the hearing threshold sound. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 35 = 10 log(I/Io) I/Io = 10^3.5 I = 3162.28 Io approx. 83 = 10 log(I/Io) I/Io = 10^8.3 I = 199526231.5 Io 117 = 10 log(I/Io) I/Io = 10^11.7 I = 5.011872336 * 10^11 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** the equation to find I is dB = 10 log(I / I0) so the equations would be 35 = 10 log(I / I0) 83 = 10 log(I / I0) 117 = 10 log(I / I0). The solution for I in the equation dB = 10 log(I / I0) is I = I0 * 10^(dB/10). For the given values we would get solutions 10^(35/10) I0 = 3162.3 I0 10^(83/10) I0 = 199526231.5 I0 10^(117/10) I0 = 501187233627 I0 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q9. is log(x^y) = x log(y) valid? If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, the x and y are reversed, should be y log(x) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x/y) = log(x) - log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, this is valid from the table confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aYes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y). STUDENT COMMENT I can verify these just fine if I have a copy of the laws in front of me; but I don't see any way to remember them all. INSTRUCTOR RESPONSE Let's look at a way to understand the laws. log( x * y ) = log(x) + log(y). For example if x = 1000 and y = 100: x * y = 1000 * 100 = 10^3 * 10^2 = 10^(3 + 2) = 10^5 so log(x * y) = log(10^5) = 5. log(x) = log(100) = log(10^2) = 2, so log(x) = 2. log(y) = log(1000) = log(10^3) = 3, so log(y) = 3. log(x) + log(y) = 2 * 3 = 5, and log(x * y) = 5. So for this example we have verified that log(x * y) = log(x) + log(y). If we think about it we also see why this must be true. When we multiply x * y the exponents of these numbers add. Since the log is the exponent, the logs therefore add. log(x / y) = log(x) - log(y). For example if x = 1000 and y = 100: x / y = 1000 / 100 = 10^3 / 10^2 = 10^(3 - 2) = 10^1, so log(x / y) = log(10^1) = 1. log(x) - log(y) = 3 - 2 = 1. This verifies the law, and shows how it is connected to the law of exponents for division. log(x^a) = a log(x). For example if x = 1000 and a = 5: log(x^a) = log(1000 ^ 5) = log( (10^3)^5 ) = log(10^15) = 15. a log(x) = 5 * log(10^3) = 5 * 3 = 15. This verifies the law and shows how it is connected to the law of exponents for powers of a number. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log (x * y) = log(x) * log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: NO, it should be log(x) + log(y) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x * y) = log(x) + log(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis 2 log(x) = log(2x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, it should be log(x^2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x + y) = log(x) + log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, First part should be log(x * y) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(x) + log(y) = log(xy), not log(x+y). ** STUDENT QUESTION: I understand that log(x) + log(y) isn't log(x + y), because the rule says that log(x) + log(y) = log(x y). Still it looks like log(x) + log(x) should be log(x + y), and I'm having trouble keeping this straight. INSTRUCTOR RESPONSE: Another way to see that it's not so is to use numbers. log(1000) = 3 and log(100) = 2, so log(1000) + log(100) = 3 + 2 = 5. However log(1000 + 100) = log(1100). Since 1100 is between 10^3 = 1000 and 10^4 = 10 000, log(1100) is between 3 and 4. (Also since 1100 is a whole lot closer to 1000 than to 10 000, we know that log(1100) is a lot closer to 3 than to 4). In any case, log(1000 + 100) certainly isn't 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x) + log(y) = log(xy) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, it is the inverse to a^x * a^y = a^(x+y) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThis is value. It is inverse to the law of exponents a^x*a^y = a^(x+y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x^y) = (log(x)) ^ y valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, should be y log(x) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x - y) = log(x) - log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, log(x/y) = log(x) - log(y) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x-y) = log x/ log y &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### The table states log(a/b) = log(a) - log(b)#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis 3 log(x) = log(x^3) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, Table states b log(a) = log(a^b) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aYes. log(x^a) = a log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x^y) = y + log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, log(x^y) = y log(x) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x^y) = y log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x/y) = log(x) / log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No, log(x/y) = log(x) - log(y) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x/y) = log(x) - log(y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qis log(x^y) = y log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, log(a^b) = b log(a) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThis is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q10. what do you get when you simplify log {base 8} (1024)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(1024) / log(8) = approx. 3.33, 3 is continuous, so the answer is 3 1/3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aCOMMON ERROR: log {base 8} (1024) = Log (1024) / Log (8) = 3.33333 EXPLANATION: log {base 8} (1024) = Log (1024) / Log (8) is correct, but 3.33333 is not an exact answer. log {base 8 } (1024) = log {base 8 } (2^10). Since 8 = 2^3, 2^10 = 2^(3 * 10/3) = (2^3)^(10/3) = 8^(10/3). Thus log {base 8} 1024 = log{base 8} 8^(10/3) = 10/3. Note that 10/3 is not exactly equal to 3.33333. You need to give exact answers where possible. STUDENT QUESTION I recognized from computer terms that 8 and 1024 were related, but I wasn't sure how they could be brought to the same level so to speak for the logs to be evaluated into fractions. I think this is the confusing bit for me: 2^10 = 2^(3 * 10/3). I see that 3*10/3 = 10, but I'm not sure how you arrived at making this relationship. Could you provide a little more explanation? INSTRUCTOR RESPONSE We want to find what power of 8 is equal to 1024. We know that 8 = 2^3 and 1024 = 2^10. So the question becomes: what power of 2^3 is 2^10? We can reason out the answer as indicated, but I would agree if you said that isn't something that will naturally occur to most students. Alternatively we can let p be the power we're looking for and write (2^3)^p = 2^10. We easily enough solve this by applying the laws of exponents. Since (2^3)^p = 2^(3 p) we get 2^(3 p) = 2^10 so that 3 p = 10 and p = 10/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat do you get when you simplify log {base 2} (4 * 32)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(4*32) / log(2) = 7 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### I understand that 4 is same as 2^2 and 32 is 2^5 and when simplified the 2s cancels out in bases leaving (2 + 5) in the powers to equal 7#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat do you get when you simplify log (1000)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There are 3 zeros in 1000 so log(1000) = 3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSince 10^3 = 1000, we have log (1000) = 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat do you get when you simplify ln(3xy)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By properties of logs ln(3xy) = ln(3) + ln(x) + ln(y) = 1.099 + ln(x) + ln(y) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat do you get when you simplify log(3) + log(7) + log(41)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(3 * 7 * 41) = 2.935 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a log(3) + log(7) + log(41) = log (3*7*41). 3 * 7 * 41 is not a rational-number power of 10 so this can't be evaluated exactly. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####I understand that 3, 7, 41 or the product of these are not a power of 10 in order to get an exact number#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q11. Show how you used the given values to find the logarithm of 12. Explain why the given values don't help much if you want the log of 17. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log(12) = log(2 * 2 * 3) = log(2) + log(2) = log(3) = .301 + .301 + .477 = 1.079 The given values dont help because no combination of products equals 17 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The problem was as follows: Given that log(2) = .301, log(3) = .477, log(5) = .699, log(7) = .845, find log(4), log(6), log(8), and log(9). Use the given values to find the logarithm of every possible integer between 11 and 20. To get log(12), given the logs of 2, 3, 5 and 7, you have to break 12 down into a product of these numbers. Since 12 = 2 * 2 * 3 we have log(12) = log(2) + log(2) + log(3) = .301 + .301 + .477 = 1.079. Your calculator will confirm this result. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x log(3) = (x - 4) log(7) 2x log(3) = x log(7) - 4 log(7) 2x(.477) = .845x - 3.3803 .9542x = .845x - 3.3803 .1091x = -3.3803 x = -30.972 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a ** log[3^(2x)]= log [7^(x-4)]. Using the laws of logarithms we get 2xlog(3)= (x-4) log(7). The distributive law gives us 2xlog(3)= xlog(7)- 4log(7). Rearranging to get all x terms on one side we get 2xlog(3)- xlog(7)= -4log(7). Factor x out of the left-hand side to get x ( 2 log(3) - log(7) ) = -4 log(7) so that x = -4 log(7) / [ 2 log(3) - log(7) ]. Evaluating this we get x = -31, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### my process was different than given answer, but I solved for the logs and used algebra to find same answer#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure how to solve using properties of logs, but by plugging in random numbers I got x = .61 approx. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aCOMMON ERROR: 3xlog(2) + 4xlog(2) = 9 Explanation: Your equation would require that log( 2^(3x) + 2^(4x) ) = log(2^(3x) ) + log(2^(4x)). This isn't the case. log(a + b) is not equal to log(a) + log(b). log(a) + log(b) = log(a * b), not log(a + b). If this step was valid you would have a good solution. However it turns out that this equation cannot be solved exactly for x. The best we can do is certain sophisticated forms of trial and error. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): #### I see why I was confused, but is the only way to solve these type of equations is by trial and error????? ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat do you get when you solve 3^(2x-1) * 3^(3x+2) = 12, and how did you solve the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3^(5x + 1) = 12 5xlog(3) + 1log(3) = log(12) 5xlog(3) + .4771 = 1.0792 5xlog(3) = .6021 x = .6021 / (5log(3)) x = .2524 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** 3^(2x-1) * 3^(3x+2) = 12. Take log of both sides: log{3} [3^(2x-1) * 3^(3x+2)] = log{3} 12. Use log(a*b) = log(a) + log(b): log{3}(3^(2x-1)) + log{3}(3^(3x+2) = log{3} 12. Use laws of logs: (2x-1) + (3x+2) = log{3} 12. Rearrange the left-hand side: 5x + 1 = log{3}12. Subtract 1 from both sides then divide both sides by 5: x = (log {3}(12) -1)/ 5. Use the fact that log{b}(x) = log x / (log b) to get x = (log(12) / log(3) - 1) / 5. Evaluate using calculator: x = .2524 ** Alternate solution: log[3^(2x-1) * 3^(3x+2)] = log (12)
.............................................
Given Solution: `a** Recall when we substituted the coordinates of three points into the form y = a x^2 + b x + c of a quadratic, then solved the resulting system of three linear equations to get the values of a, b and c. We do something similar here, but this time the form is y = A * 2^(kx). Our two unknowns will be A and k, and we will only require two equations to solve for our two unknowns. We therefore need only two points. Substituting data points into the form y = A * 2^(kx) we get 3 = A * 2^(-4k) (from the data point (-4,3)) and 2 = A * 2^( 7k) (from the data point (7,2)). When we solved the three equations for the quadratic model, we added multiples of equations in order to eliminate variables. However adding multiples of equations doesn't work here. What does work is dividing one equation by another, which eliminates the variable A. Dividing the first equation by the second we get 3/2 = A * 2^(-4k) / (A * 2^(7k) ), which simplifies to 1.5 = 2^(-4k)/ 2^(7k). Now by the laws of exponents 2^(-4k)/ 2^(7k) = 2^(-4-7k)= 2^(-11 k), so that log(2^(-11 k)) = log(1.5) and -11 k * log(2) = log 1.5. We solve this for k, obtaining k = log(1.5) / (-11 log(2)). Evaluating with a calculator we find that k = -.053, approx.. Just as we did when solving for the quadratic model, once we have evaluated an unknown we substitute that value into our original equations. Our first equation was 3 = A * 2^(-4k), which is easily solve for A to give us A = 3 / (2 ^(-4k) ). Substituting k = -.053 we get A= 3/ 1.158 = 2.591. Now we know the values of both A and k, which we substitute into our form y = A * 2^(kx) to obtain our model y = 2.591(2^-.053t). ** STUDENT COMMENT: I'm trying, but I just don't understand the given solution. INSTRUCTOR RESPONSE: You've understood just about everything up to this point. Here's an expanded solution, with some questions you should consider. You are welcome to submit a copy of this part and insert your responses, indicated by #$&* before and after each insertion: Recall when we substituted the coordinates of three points into the form y = a x^2 + b x + c of a quadratic, then solved the resulting system of three linear equations to get the values of a, b and c. We do something similar here, but this time the form is y = A * 2^(kx). Our two unknowns will be A and k, and we will only require two equations to solve for our two unknowns. We therefore need only two points. Substituting data points into the form y = A * 2^(kx) we get 3 = A * 2^(-4k) (from the data point (-4,3)) and 2 = A * 2^( 7k) (from the data point (7,2)). Do you understand how the given points and the form y = A * 2^(k x) leads to these equations? When we solved the three equations for the quadratic model, we added multiples of equations in order to eliminate variables. However adding multiples of equations doesn't work here. What does work is dividing one equation by another, which eliminates the variable A. Dividing the first equation by the second we get 3/2 = A * 2^(-4k) / (A * 2^(7k) ), which simplifies to 1.5 = 2^(-4k)/ 2^(7k). Do you understand how dividing the equation 3 = A * 2^(-4k) by the equation 2 = A * 2^( 7k) leads to the equation 1.5 = 2^(-4k)/ 2^(7k)? Now by the laws of exponents 2^(-4k)/ 2^(7k) = 2^(-4-7k)= 2^(-11 k), so that log(2^(-11 k)) = log(1.5) and -11 k * log(2) = log 1.5. We solve this for k, obtaining k = log(1.5) / (-11 log(2)). Do you understand how the right-hand side simplifies to 2^(-11 k), using the laws of exponents? Evaluating with a calculator we find that k = -.053, approx.. Just as we did when solving for the quadratic model, once we have evaluated an unknown we substitute that value into our original equations. Our first equation was 3= A * 2^(-4k), which we easily solve for A to get A = 3 / (2 ^(-4k) ). Substituting k = -.053 we get A= 3/ 1.158 = 2.591. Do you understand how substituting k = -.053 into the first equation leads to A = 2.591? Now we know the values of both A and k, which we substitute into our form y = A * 2^(kx) to obtain our model y = 2.591(2^-.053t). Do you understand that this is our original form y = A ( 2 * (k t) ), where A and k have been replaced by the values we obtained by solving the simultaneous equations? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ####Very interesting, it did not occur to me to divide the two equations together in order to eliminate the A value, I got hung up on this crucial step, but I understand this and how to manipulate the rest of the problem by using law of exponents first to get 2^-11k = 1.5, then it just a matter of taking the log of both sides, -11klog(2) = log(1.5) and then dividing both sides by -11log(2) to find the k value, then plugging the k value into one equation to solve for A.#### ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat is the exponential function of form A e^(k2 t) such that the graph passes thru points thru points (-4,3) and (7,2) and how did you solve the equations to find this function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 = A * e^(-4k) 2 = A * e^(7k) 1.5 = (e^(-4k)) / (e^(7k)) 1.5 = e^(-11k) ln is inverse to e ln(1.5) = -11k k = -.0369 plug in to get 2 = A * e^(7*-.0369) 2 = A * .7724 A = 2.589 y = 2.589 * e^(-.0369t) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Substituting data points into the form y = A * e^(kx) we get 3= A * e^(-4k) and 2= A * e^(7k) Dividing the first equation by the second we get 1.5= e^(-4k)/ e^(7k)= e^(-4-7k)= 2^(-11k) so that ln(e^(-11k)) = ln(1.5) and -11 k = 1.5 so that k= ln(1.5) / (-11). Evaluating with a calculator: k= -.037 approx. From the first equation A = 3 / (e ^(-4k) ). Substituting k = -.037 we get A= 3/ 1.158 = 2.591. So our form y = A * e^(kx) gives us y= 2.591(e^-.039 t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat is the exponential function of form A b^t such that the graph passes thru points thru points (-4,3) and (7,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 = A * b^-4 2 = A * b^7 1.5 = b^-11 b = .9639 approx. 2 = A * .9639^7 A = 2.587 y = 2.587 * .9639^t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Our equations are 3= Ab^-4 2= Ab^7 3/2= Ab^-4/(Ab^7) 1.5= b^-11 b= .96 3= A * .96 ^ -4 3= A * 1.177 2.549= A y= 2.549 * .96^t ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: ********************************************* Question: `q2. Find the exponential function corresponding to the points (5,3) and (10,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 = A * b^5 2 = A * b^10 1.5 = b^-5 b = .9231 approx. 3 = A * .9231^5 A = 4.476 approx. Equation is y = 4.476 * .9231^t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Using y = A b^t we get equations 3= Ab^5 2= Ab^10 Dividing first by second: 3/2= Ab^5/(Ab^10). 1.5= b^-5 b= .922 Now A = 3 / b^5 = 3 / .922^5 = 4.5. Our model is y = 4.5 * .922^t. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat are k1 and k2 such that b = e^k2 = 2^k1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln is inverse of e so using given b in previous solution ln(.922) = k2, where k2 = -.0812 approx. .922 = 2^k1 k 1 = -.117 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** .922 = e^k2 is directly solved by taking the natural log of both sides to get k2 = ln(.922) = -.081. .922= 2^k1 is solved as follows: log(.922) = log(2) k1 k1 = log(.922) / log(2) = -.117 approx.. Using these values for k1 and k2 we get }g(x) = A * 2^(k1 t) = 4.5 * 2^(-.117 t) and h(x) = A e^(k2 t) = 4.5 e^(-.081 t). **** STUDENT QUESTION I understand the process but I'm a little lost as to what b = e^k2 = 2^k1 is supposed to tell us. What are k1 and k2 and how are they related? INSTRUCTOR RESPONSE The form A e^(kx) is pretty standard for exponential functions. The main reason is that it's very easy to do calculus when the base is e; that's the most natural base to use for calculating rate-of-change functions; and that's why its inverse function is called the 'natural log'). The form A * 2^(kx) is useful because in this form we can easily calculate doubling time and/or halflife (using this form it's not hard to see why doubling time is the value of x such that k x = 1, halflife is the value of x such that kx = -1). To get the relationship between k_2 and k_1: Starting with e^k_2 = 2^k_1 , take the natural log of both sides to get k_2 = ln(2^(k_1)) = k_1 ln(2). Thus k_2 = k_1 ln(2), and that's the relationship between k_1 and k_2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q3. earthquakes measure R1 = 7.4 and R2 = 8.2. What is the ratio I2 / I1 of intensity and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (Io * 10^R2) / (Io * 10^R1) = 10^(R2 - R1) = 10^(8.2 - 7.4) = 10^.8 = 6.310 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** R1 = log(I1 / I0) and R2 = log(I2 / I0) so I1/I0 = 10^R1 and I1 = 10^R1 * I0 and I2/I0 = 10^R2 and I2 = 10^R2 * I0 so I2 / I1 = (I0 * 10^R2) / (I0 * 10^R1) = 10^R2 / 10^R1 = 10^(R2-R1). So if R2 = 8.2 and R1 = 7.4 we have I2 / I1 = 10^(R2 - R1) = 10^(8.2 - 7.4) = 10^.8 = 6.3 approx. An earthquake with R = 8.2 is about 6.3 times as intense as an earthquake with R = 7.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q4. I2 / I1 ratios If one earthquake as an R value 1.6 higher than another, what is the ratio I2 / I1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10^1.6 = 39.811 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** As before I2 / I1 = 10^(R2-R1). If R2 is 1.6 greater than R1 we have R2 - R1 = 1.6 and I2 / I1 = 10^1.6 = 40 approx. An earthquake with R value 1.6 higher than another is 40 times as intense. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qIf one earthquake as an R value `dR higher than another, what is the ratio I2 / I1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10^dR confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** As before I2 / I1 = 10^(R2-R1). If R2 is `dR greater than R1 we have R2 - R1 = `dR and I2 / I1 = 10^`dR. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating:3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*
#$&* course now find synopsis of ph1 etc from pictures ... probably in ph2 of shell_01 on netbook ... or on server ...
.............................................
Given Solution: During the first year the interest will be 10% of $1000, or $100. This makes the total at the end of the first year $1100. During the second year the interest will be 10% of $1100, or $110. At the end of the second year the total will therefore be $1100 + $110 = $1210. During the third year the interest will be 10% of $1210, or $121. At the end of the sphere year the total will therefore be $1210 + $121 = $1331. The yearly changes are $100, $110, and $121. These changes increase year by year. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. In the preceding problem you obtained amounts $1100, $1210 and $1331. What number would you multiply by $1000 to get $1100? What number we do multiply by $1100 to get $1210? What number would we multiply by $1210 to get $1331? What is the significance of this number and how could we have found it from the original information that the amount increases by 10 percent each year? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used the formula y = 1000 * 1.1^t, so multiplied by 1.1, this number taken to power time multiplied by initial gives answer. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: To get $1100 you have to multiply $1000 by 1100 / 1000 = 1.1. To get $1210 you have to multiply $1210 by 1210 / 1100 = 1.1. To get $1331 you have to multiply $1331 by 1331 / 1210 = 1.1. If the amount increases by 10 percent, then you end up with 110 percent of what you start with. 110% is the same as 1.1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q003. Given the recurrence relation P(n) = 1.10 * P(n-1) with P(0) = 1000, substitute n = 1, 2, and 3 in turn to determine P(1), P(2) and P(3). How is this equation related to the situation of the preceding two problems? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(1) = 1.1 * P(0) = 1100 P(2) = 1.1 * P(1) = 1210 P(3) = 1.1 * P(2) = 1331 This equation gives the answer by multiplying 1.1 by the previous years total/ confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Substituting 1 into P(n) = 1.10 * P(n-1) we obtain P(1) = 1.10 * P(1-1), or P(1) = 1.10 * P(0). Since P(0) = 1000 we get P(1) = 1.1 * 1000 = 1100. Substituting 2 into P(n) = 1.10 * P(n-1) we obtain P(2) = 1.10 * P(1). Since P(1) = 1100 we get P(1) = 1.1 * 1100 = 1210. Substituting 3 into P(n) = 1.10 * P(n-1) we obtain P(3) = 1.10 * P(2). Since P(2) = 1000 we get P(3) = 1.1 * 1210 = 1331. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q004. If you are given $5000 and invest it at 8% annual interest, compounded annually, what number would you multiply by $5000 to get the amount at the end of the first year? Using the same multiplier, find the results that the end of the second and third years. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By 1.08 equation is y = 5000 * 1.08^t 1st year is $5400 2nd is $ 5832 3rd year is $6298.56 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If your money increases by 8% in a year, then it the end of the year you will have 108% as much as at the beginning. Since 108% is the same as 1.08, our yearly multiplier will be 1.08. If we multiply $5000 by 1.08, we obtain $5000 * 1.08 = $5400, which is the amount the end of the first your. At the end of the second year the amount will be $5400 * 1.08 = $5832. At the end of the third year the amount will be $5832 * 1.08 = $6298.56. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q005. How would you write the recurrence relation for a $5000 investment at 8 percent annual interest? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: P(n) = 1.08 * P(n -1) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Just as the recurrence relation for 10 percent annual interest, as seen in the problem before the last, was P(n) = 1.10 * P(n-1), the recurrence relation for 8 percent annual interest is P(n) = 1.08 * P(n-1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. If you are given amount $5000 and invest it at annual rate 8% or .08, then after n years how much money do you have? What does a graph of amount of money vs. number of years look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: money = $5000 * 1.08^n y-intercept is (0, 5000) graph is exponential increasing at an increasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: After 1 year the amount it $5000 * 1.08. Multiplying this by 1.08 we obtain for the amount at the end of the second year ($5000 * 1.08) * 1.08 = $5000 * 1.08^2. Multiplying this by 1.08 we obtain for the amount at the end of the third year ($5000 * 1.08^2) * 1.08 = $5000 * 1.08^3. Continuing to multiply by 1.08 we obtain $5000 * 1.08^3 at the end of year 3, $5000 * 1.08^4 at the end of year 4, etc.. It should be clear that we can express the amount at the end of the nth year as $5000 * 1.08^n. If we evaluate $5000 * 1.08^n for n = 0, 1, 2, ..., 10 we get $5000, $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62. It is clear that the amount increases by more and more with every successive year. This result in a graph which passes through the vertical axis at (0, 5000) and increases at an increasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q007. With a $5000 investment at 8 percent annual interest, how many years will it take to double the investment? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 10000 = 5000 * 1.08^n 2 = 1.08^n n = 9.00646 approx. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Multiplying $5000 successively by 1.08 we obtain amounts $5400, 5832, 6298.56, 6802.45, 7346.64, 7934.37, 8569.12, 9254.65, 9995.02, 10,794.62 at the end of years 1 thru 10. We see that the doubling to $10,000 occurs very shortly after the end of the ninth year. We can make a closer estimate. If we calculate $5000 * 1.08^x for x = 9 and x = 9.1 we get about $10,072. So at x = 9 and at x = 9.1 the amounts are $9995 and $10072. The first $5 of the $77 increase will occur at about 5/77 of the .1 year time interval. Since 5/77 * .1 = .0065, a good estimate would be that the doubling time is 9.0065 years. If we evaluate $5000 * 1.08^9.0065 we get $10,000.02. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q008. If you are given amount P0 and invest it at annual rate r (e.g., for the preceding example r would be 8%, which in numerical form is .08), then after n years how much money do you have? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = P0 * 1.08^n confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If the annual interest rate is .08 then each year we would multiply the amount by 1.08, the amount after n years would be P0 * 1.08^n. If the rate is represented by r then each year then each year we multiply by 1 + r, and after n years we have P0 * (1 + r)^n. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q009. If after an injection of 800 mg an antibiotic your body removes 10% every hour, then how much antibiotic remains after each of the first 3 hours? How long does it take your body to remove half of the antibiotic? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 800 * .1^n 800 * .9^1 = 720mg left 800 * .9^2 = 648mg left 800 * .9^3 = 583.2 mg left For body to remove half takes about 6.61 approx hours. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: If 10 percent of the antibiotic is removed each hour, then at the end of the hour the amount left will be 90 percent of what was present at the beginning of the hour. Thus after 1 hour we have .90 * 800 mg, after a second hour we have .90 of this, or .90^2 * 800 mg, and after a third hour we have .90 of this, or .90^3 * 800 mg. The numbers are 800 mg * .90 = 720 mg, then .90 of this or 648 mg, then .90 of this or 583.2 mg. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q010. In the preceding problem, what function Q(t) represents the amount of antibiotic present after t hours? What does a graph of Q(t) vs. t look like? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Q(t) = 800 * .9^t y-intercept is (0, 800) graph decreases at decreasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: After t hours we will have 800 * .9^t mg left. So Q(t) = 800 * .9^t. The amounts for the first several years are 800, 720, 648, 583.2, etc.. These amounts decrease by less and less each time. As a result the graph, which passes through the vertical axes at (0,800), decreases at a decreasing rate. We note that no matter how many times we multiply by .9 our result will always be greater than 0, so the graph will keep decreasing at a decreasing rate, approaching the horizontal axis but never touching it. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q011. Suppose that we know that the population of fish in a pond, during the year after the pond is stocked, should be an exponential function of the form P = P0 * b^t, where t stands for the number of months after stocking. If we know that the population is 300 at the end of 2 months and 500 at the end of six months, then what system of 2 simultaneous linear equations do we get by substituting this information into the form of the function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 300 = P0 * b^2 500 = P0 * b^6 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: We substitute the populations 300 then 500 for P and substitute 2 months and 6 months for t to obtain the equations 300 = P0 * b^2 and 500 = P0 * b^6. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q012. We obtain the system 300 = P0 * b^2 500 = P0 * b^6 in the situation of the preceding problem. If we divide the second equation by the first, what equation do we obtain? What do we get when we solve this equation for b? If we substitute this value of b into the first equation, what equation do we get? If we solve this equation for P0 what do we get? What therefore is our specific P = P0 * b^t function for this problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 5/3 = 1 * b^4 b = 1.136 approx. 300 = P0 * 1.136^2 P0 = 232.469 approx. P = 232.469 * 1.136^t confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: Dividing the second equation by the first the left-hand side will be left-hand side: 500/300, which reduces to 5/3, and the right-hand side will be right-hand side: (P0 * b^6) / (P0 * b^2), which we rearrange to get (P0 / P0) * (b^6 / b^2) = 1 * b^(6-2) = b^4. Our equation is therefore b^4 = 5/3. To solve this equation for b we take the 1/4 power of both sides to obtain (b^4)^(1/4) = (5/3)^(1/4), or b = 1.136, to four significant figures. Substituting this value back into the first equation we obtain 300 = P0 * 1.136^2. Solving this equation for P0 we divide both sides by 1.136^2 to obtain P0 = 300 / (1.136^2) = 232.4, again accurate to 4 significant figures. Substituting our values of P0 and b into the original form P = P0 * b^t we obtain our function P = 232.4 * 1.136^t. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating:3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*
#$&* course cal1 larson for modificationgiancoli text goals (and lots of other stuff)