#$&* course mth 163 7/4 9am 017. `q uery 17*********************************************
.............................................
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: ********************************************* Question: `qfor what value of x will the function y = ln(x) first reach y = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is the inverse of e (x) a table shows x = 4 and y = 54.598 So flip these to get x = 54.598 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: ********************************************* 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: Because the negative x axis is an asymptote of b^x and this is the inverse of that. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: For the 1st one (1, 3.5) so b=3.5 For the 2nd one (1, 7.1) so b=7.1 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qAt what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: For 7.1^x: y=2, x=.33 y=3, x=.55 y=4, x=.66 For 3.5^x: y=2, x= .66 y=3, x= .9 y=4, x= 1.125 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** Using trial and error we find that for y = 3.5^x, the x values .55, .88 and 1.11 give us y = 2, 3 and 4. We find that for y = 7.3^x, the x values .35, .55 and .70 give us y = 2, 3 and 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The threshold intensity is 10000 Log(10000) = 4, so db = 10*4 = 40 this is the decibel level. confidence rating #$&*: 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): ------------------------------------------------ Self-critique rating: ********************************************* 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, so 10*2 = 20 Log (10000000) = 7, so 10*7 = 70 Log (1000000000) = 9, so 10*9 = 90 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a10 log(100) = 10 * 2 = 20, so a sound which is 100 times hearing threshold intensity is a 20 dB sound. 10 log(10,000,000) = 10 * 7 = 70, so a sound which is 100 times hearing threshold intensity is a 70 dB sound. 10 log(1,000,000,000) = 10 * 9 = 90, so a sound which is 100 times hearing threshold intensity is a 90 dB sound. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qhow can you easily find these decibel levels without using a calculator? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You could use 10 to a certain power because a log is the inverse of a power. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique rating: ********************************************* 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: Log (500) = 2.6989, 10*2.6989 = 26.9897db Log (30000000) = 7.4771, 10*7.4771 = 74.7712db Log (7000000000) = 9.845, 10*9.845 = 98.45db confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a10 log(500) = 10 * 2.699 = 26.99 so this is a 26.99 dB sound. 10 log(30,000,000) = 10 * 7.477 = 74.77 so this is a 74.77 dB sound. 10 log(7,000,000,000) = 10 * 9.845 = 98.45 so this is a 98.45 dB sound. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would first divide 40 by 10 to get 4, then we have log (x) = 4 the inverse of this would be 10^4 = 10000 is the answer confidence rating #$&*: 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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qAnswer the same question for sounds measuring 20, 50, 80 and 100 decibels. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 20/10 = 2 which leaves log(x) = 2 then 10^2 = 100 50/10 = 5 which leaves log(x) = 5 then 10^5 = 100,000 80/10 = 8 which leaves log (x) = 8 then 10^8 = 100,000,000 100/10 = 10 which leaves log (x) = 10 then 10^10 = 10,000,000,000 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** since dB = 10 log(I / I0) we have log(I/I0) = dB / 10. Translating to exponential form this tells us that I / I0 = 10^(dB/10) wo that I = I0 * 10^(dB/10). For a 20 dB sound this gives us I = I0 * 10^(20/10) = I0 * 10^2 = 100 I0. The sound is 100 times the intensity of the hearing threshold sound. For a 50 dB sound this gives us I = I0 * 10^(50/10) = I0 * 10^5 = 100,000 I0. The sound is 100,000 times the intensity of the hearing threshold sound. For an 80 dB sound this gives us }I = I0 * 10^(80/10) = I0 * 10^8 = 100,000,000 I0. The sound is 100,000,000 times the intensity of the hearing threshold sound. For a 100 dB sound this gives us I = I0 * 10^(100/10) = I0 * 10^10 = 10,000,000,000 I0. The sound is 10,000,000,000 times the intensity of the hearing threshold sound. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: db = 10*log (I/I0) 35=10*log(I/I0) = 10^(35/10) = 3162.277 83=10*log(I/Io) = 10^(83/10) = 199526231.5 177=10*log(I/I0) = 10^(117/10) = 501187233627 confidence rating #$&*: 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): ------------------------------------------------ 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 it is not valid because 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: