#$&* course Mth163 7/8/2013 around 14:00
.............................................
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): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qfor what value of x will the function y = ln(x) first reach y = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&& The inverse of the function Ln(x) (natural log) is e^x. e^4=54.6 (approx.) to check I just plug this backinto ln(54.6)=4 (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): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q3. Explain why the y axis is an asymptote for a log{base b}(x) function explain why the asymptote is the negative y axis if and only if b > 1 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& It approaches an asymptote it keeps getting smaller and smaller but never actually going past the asymptotes number
.............................................
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 can’t 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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q5. What are your estimates for the values of b for the two exponential functions on the given graph? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& The graphs y int is through (0,1) and b is the y intercept &&&& confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The exponential function y = A b^x has y intercept (0, A) and basic point (1, A * b). Both graphs pass thru (0, 1) so A = 1. The x = 1 points are (1, 3.5) and (1, 7.3) approx.. Thus for the first A * b = 3.5; since A = 1 we have b = 3.5. For the second we similarly conclude that b = 7.3. So the functions are y = 3.5^x and y = 7.3^x, approx.. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& When you say a=1 you are taking about the constant slope right? The differences between the change in ys and change in x values I am not really sure how to apply slope with imaginary numbers as I had gotten in my results I also feel that those are probably pretty wrong
.............................................
Given Solution: `a** Using trial and error we find that for y = 3.5^x, the x values .55, .88 and 1.11 give us y = 2, 3 and 4. We find that for y = 7.3^x, the x values .35, .55 and .70 give us y = 2, 3 and 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&&& “ Question: `qAt what points will each of the*logarithmic* functions reach the values y = 2, y = 3 and y = 4?” &&&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& Db=10*log(I/I0)=10*log(10,000/1,000,000,000,000)=-80 &&&& confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Why would we be trying to find the inverse???
.............................................
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): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qhow can you easily find these decibel levels without using a calculator? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& 10 * 2 = 20 10 * 7 = 70 10 * 9 = 90 Just take the inverse and multiply the 10 by the number of zeros in your number &&&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSince 10^1, 10^2, 10^3, ... are 10, 100, 1000, ..., the power of 10 is the number of zeros in the result. Since the log of a number is the power to which 10 must be raised to get this number, the log of one of these numbers is equal to the number of zeros.
.............................................
Given Solution: `a10 log(500) = 10 * 2.699 = 26.99 so this is a 26.99 dB sound. 10 log(30,000,000) = 10 * 7.477 = 74.77 so this is a 74.77 dB sound. 10 log(7,000,000,000) = 10 * 9.845 = 98.45 so this is a 98.45 dB sound. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& We still keep the 10 where it is and take the log of the actual number first then times 10 by the outcome of log and intial number 10 log(500) = 10 * 2.699 = 26.99 10 log(30,000,000) = 10 * 7.477 = 74.77 10 log(7,000,000,000) = 10 * 9.845 = 98.45 This makes sense &&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& Knowing that 40 dB=10^4=10,000 this is our intensity of sound &&& confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 understand how to do the rest but I = 10,000 I0 when you group these two in the beginning saying treat them as x, how do they separate??? &&&& ------------------------------------------------ Self-critique rating:
.............................................
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): &&&& Okay this explains what is going on with the x &&&& ------------------------------------------------ Self-critique rating:OK
.............................................
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): &&& My cal just gave me so many numbers then a small cap E with what I presume was the number of decimals so I just filled in the numbers I did not know with zeros &&&& ------------------------------------------------ Self-critique rating: ********************************************* Question: `q9. is log(x^y) = x log(y) valid? If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& No, because the rules of logs say that log(a^b)=b log (a) instead of log(x^y) = x log(y) it would be log(x^y) = y log(x) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x/y) = log(x) - log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Yes, this is in agreement with the rules of logs a^x / a^y = a^(x-y) &&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log (x * y) = log(x) * log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& No , the right way is (x * y) = log(x) +log(y) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x * y) = log(x) + log(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis 2 log(x) = log(2x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&& No, it is supposed to be raised to the x, 2 log(x) = log(2^x) &&&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* 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 #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** log(x) + log(y) = log(xy), not log(x+y). ** STUDENT QUESTION: I understand that log(x) + log(y) isn't log(x + y), because the rule says that log(x) + log(y) = log(x y). Still it looks like log(x) + log(x) should be log(x + y), and I'm having trouble keeping this straight. INSTRUCTOR RESPONSE: Another way to see that it's not so is to use numbers. log(1000) = 3 and log(100) = 2, so log(1000) + log(100) = 3 + 2 = 5. However log(1000 + 100) = log(1100). Since 1100 is between 10^3 = 1000 and 10^4 = 10 000, log(1100) is between 3 and 4. (Also since 1100 is a whole lot closer to 1000 than to 10 000, we know that log(1100) is a lot closer to 3 than to 4). In any case, log(1000 + 100) certainly isn't 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x) + log(y) = log(xy) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Yes it agrees with the rules log(a) + log(b) = log(ab) &&&& confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThis is value. It is inverse to the law of exponents a^x*a^y = a^(x+y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& Inverse to LOE a^(x)*a^(y)=(x+y) &&& ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x^y) = (log(x)) ^ y valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& No it should be log(x^y) =y log(x) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x - y) = log(x) - log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& No it should be log(x /y) = log(x) - log(y) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x-y) = log x/ log y &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& I’m pretty sure it is supposed to be log(x /y) = log(x) - log(y) that is what you have on your notes anyway Labeled Logarithms, logarithmic functions, logarithmic equations &&&&
.............................................
Given Solution: `aYes. log(x^a) = a log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x^y) = y + log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& No plus sign , take that out and its good log(x^y) = y log(x) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x^y) = y log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x/y) = log(x) / log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& NO log(x/y) = log(x) - log(y) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aNo. log(x/y) = log(x) - log(y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qis log(x^y) = y log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& Yes &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThis is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK
.............................................
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): ------------------------------------------------ Self-critique rating:OK ********************************************* 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{base 2}(4*32) = [log(4)/log(2)]+[log(32)/log(2)]=2+5=7 7
.............................................
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): &&& Will this way come in handy later, because it seems a little more complicated to me. `a** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. ** Is this the preferred way to do it??? &&&& ------------------------------------------------ Self-critique rating:OK ********************************************* 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: &&&& 10^3= 1000,log would be how many zeros after your one=3 &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aSince 10^3 = 1000, we have log (1000) = 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* 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: &&& ln(3xy) = ln (3) + ln(x) + ln(y) &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& ln (3)=1.0986 =1.0986+ ln(x) + ln(y) &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qwhat do you get when you simplify log(3) + log(7) + log(41)? If it can be evaluated exactly, what is the result and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& 3^x+7^x+4^x=? Using the cal my answer would be 2.935 &&& confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): &&& When logs are not rational, powe r of 10 they can not be evaluated exact. &&&& ------------------------------------------------ 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: &&& 4*3=12 We are not givien a four but we can multiyply 2 twice So log(2)+log(2)+log(3)=log(12) .301 + .301 + .477 = 1.079 &&& 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): ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 2xlog(3)=x4log(7)
.............................................
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): &&& Why wouldn’t you get all the x’s on the same side??? Oh I wrote x4 instead of x-4 and didn’t even realize it &&&
.............................................
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): ------------------------------------------------ Self-critique rating:Not given ********************************************* 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^[(2x-1)+ (3x+2)]=12 3^(5x+1)=12 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): &&& 2^(-4k)/ 2^(7k) = 2^(-4-7k)= 2^(-11 k) We can subtract because they share the same variable, otherwise it would not work, right???
.............................................
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): &&&& The problem asks us to raise to the 2k A e^(k2 t) So in the beginning why wouldn’t we multiply the 2*4 and the 2*7 while they are still powers ??? &&& ------------------------------------------------ Self-critique rating: ********************************************* 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= 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 3=2.549(.96)^-4 &&&& &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q2. Find the exponential function corresponding to the points (5,3) and (10,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 3= Ab^5 2= Ab^10 3/2= Ab^5/(Ab^10) 1.5= b^-5 After raising both sides by neg one fifth you get b which equals b= .96 A = 3 / b^5 = 3 / .922^5 = 4.5 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Using y = A b^t we get equations 3= Ab^5 2= Ab^10 3/2= Ab^5/(Ab^10). 1.5= b^-5 b= .922 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): &&& The model is y=4.5(.922)^t &&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `qWhat are k1 and k2 such that b = e^k2 = 2^k1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& 2^k1 &&& confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** .922 = e^k2 is directly solved by taking the natural log of both sides to get k2 = ln(.922) = -.081. .922= 2^k1 is solved as follows: log(.922) = log(2) k1 k1 = log(.922) / log(2) = -.117 approx.. Using these values for k1 and k2 we get }g(x) = A * 2^(k1 t) = 4.5 * 2^(-.117 t) and h(x) = A e^(k2 t) = 4.5 e^(-.081 t). **** STUDENT QUESTION I understand the process but I'm a little lost as to what b = e^k2 = 2^k1 is supposed to tell us. What are k1 and k2 and how are they related? INSTRUCTOR RESPONSE The form A e^(kx) is pretty standard for exponential functions. The main reason is that it's very easy to do calculus when the base is e; that's the most natural base to use for calculating rate-of-change functions; and that's why its inverse function is called the 'natural log'). The form A * 2^(kx) is useful because in this form we can easily calculate doubling time and/or halflife (using this form it's not hard to see why doubling time is the value of x such that k x = 1, halflife is the value of x such that kx = -1). To get the relationship between k_2 and k_1: Starting with e^k_2 = 2^k_1 , take the natural log of both sides to get k_2 = ln(2^(k_1)) = k_1 ln(2). Thus k_2 = k_1 ln(2), and that's the relationship between k_1 and k_2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): &&& k1 = log(.922) / log(2) = -.117 approx.. so .992=2^(-.117) &&&& ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q3. earthquakes measure R1 = 7.4 and R2 = 8.2. What is the ratio I2 / I1 of intensity and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 10^(8.2/7.4) =12.8256 &&& confidence rating #$&*: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): &&& Oh I though I1 stood for R1 and I2 stood for R2 Instead of dividing I should have subtracted 10^(8.2 - 7.4) = 10^.8 = 6.3 &&&& ------------------------------------------------ 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=39.8 &&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** As before I2 / I1 = 10^(R2-R1). If R2 is 1.6 greater than R1 we have R2 - R1 = 1.6 and I2 / I1 = 10^1.6 = 40 approx. An earthquake with R value 1.6 higher than another is 40 times as intense. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `qIf one earthquake as an R value `dR higher than another, what is the ratio I2 / I1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&& 10^dR=I2/I1 &&& confidence rating #$&*: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): ------------------------------------------------ Self-critique rating:OK If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: For what value of x will the value of the function y = log{base 3}(x) first reach 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&& Inverse version 3^4=81 &&&& confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: Explain why the negative y axis is an asymptote for the function y = log{base 2}(x). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&& When you take the negative powers they end up being pos fractions less than one but greater than zero They will never go past 2 So all the numbers are stuck between 0 and 2 &&&
Be sure to include the entire document, including my notes. &#
*@