course MTH 163 8:26 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** 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):Im not really sure why we used the ln(x)
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 b < 1, larger positive powers make the result smaller, and the positive x axis becomes the asymptote. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: y = 3.5^x and y = 7.3^x confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: y = 3.5^x= .55, .88 and 1.11 y = 7.3^x= .35, .55 and .70 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: log(10000)=4 10 * 4 = 40db confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 log(10,000,000)=7 log(1,000,000,000)=9 2*10=20db 7*10 = 70db 9*10 = 90db confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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)=2.699*10=26.99db log(30000000)=7.477*10=74.77db log(7000000000)=9.8451*10 =98.451db confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: log x = 4 10^4=10000 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: log x = 20=10^2=100 log x = 50=10^5=100000 log x = 80=10^8=100000000 log x= 100=10^10 = 10000000000 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 35=10 log(I/Io) 83=10 log(I/IO) 117=10 log(I/IO) confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 this is not valid log(x^y)= ylog(x) This is a property of logarithims confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* 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....this is another property of logarithims confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log (x * y) = log(x) * log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not this should be log(x) + log(y) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aNo. log(x * y) = log(x) + log(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis 2 log(x) = log(2x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: not this should be log(x)^2 not log (2x) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: `qis log(x + y) = log(x) + log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no this should be log(x)*log(y) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** log(x) + log(y) = log(xy), not log(x+y). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log(x) + log(y) = log(xy) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes this is a property of logarithims confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis is valid. It is inverse to the law of exponents a^x*a^y = a^(x+y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log(x^y) = (log(x)) ^ y valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No this should be y log (x) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: `qis log(x - y) = log(x) - log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no this should be log (x)/log (y) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aNo. log(x-y) = log x/ log y &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis 3 log(x) = log(x^3) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes this is valid confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYes. log(x^a) = a log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log(x^y) = y + log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no this should be y log(x) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aNo. log(x^y) = y log(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log(x/y) = log(x) / log(y) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: no this should be log (x) - log (y) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aNo. log(x/y) = log(x) - log(y). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qis log(x^y) = y log(x) valid. If so why, and if not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: yes this is valid confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis is valid. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: ln(1024/ln(8) = 10/3 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* 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(2)128 ln(128)/ln(2) =7 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: log(1000)/log(10) =3 confidence rating:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince 10^3 = 1000, we have log (1000) = 3 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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(3)+ln(x)+ln(y) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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) it can not be evaluated exactly confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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) + log(2) + log(3) = .301 + .301 + .477 = 1.079 12=2*2*3 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* 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)=xlog(7)-4log(7) (2x)log(3)-xlog(7)=-4log(7) x(2log(3)-log7)=-4log(7) x=-4log(7)/((2log(3)-log(7)) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Can you show me how to get the approx value ex -4 log(7)=1/2401 ? and 2log(3) = 9? and log 7 = 10^7? or am I getting this twisted up? - ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3xlog(2) + 4xlog(2)=9 0 confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 would have made this error ------------------------------------------------ Self-critique rating: ********************************************* 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: (2x-1)log(3) * (3x+2)log(3)=12 2x-1) + (3x+2)=log(3)12 5x+1=log(3)12 5x+1=2.39086-1 5x=1.39086 x=approx.278172 confidence rating:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Evaluate using calculator: x = .2524 ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qquery fitting exponential functions to data 1. what is the exponential function of form A (2^(k1 t) ) such that the graph passes thru points (-4,3) and (7,2), and what equations did you solve to obtain your result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= A *2^kx 3=A*2^k(-4) 2=A *2^k(7) 3/2=(2^k(-4))/(2^k(7)) 3/2=2^k-4-7 3/2=2^k-11 log(1.5)=log(2^-11k) -11k*log(2)=log(1.5) k=log(1.5) / (-11log(2)) k=-.053178 y=A*2^(-.053)(x) 3=A*2^(-.053)(-4) 3= A *1.15829 A=3/1.15829 A=2.59 y=2.59*2^(-.053)(x) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Substituting data points into the form y = A * 2^(kx) we get 3= A * 2^(-4k) and 2= A * 2^(7k) Dividing the first equation by the second we get 1.5= 2^(-4k)/ 2^(7k)= 2^(-4-7k)= 2^(-11k) so that log(2^(-11k)) = log(1.5) and -11 k * log(2) = log 1.5 so that k= log(1.5) / (-11log(2)). Evaluating with a calculator: k= -.053 From the first equation A = 3 / (2 ^(-4k) ). Substituting k = -.053 we get A= 3/ 1.158 = 2.591. So our form y = A * 2^(kx) gives us y= 2.591(2^-.053t). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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: y=a*e^kt (3=a*e^k(-4))/(2=A*e^k(7)) 1.5=e^(k)(-4)/e^(k)(-7) 1.5=e^k(-11) log(1.5)=log(e)(k)(-11) k=(log(1.5))/(-11log(e)) k=-.037 2=A*e(-.037)(7) 2=A*.7718 A=2.59 y=2.59*(e^(-.037)(t) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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 1.5= Ab^-4/(Ab^7) 1.5= b^-4-7 1.5=b^-11 b= .96 3= A * .96 ^ -4 3= A * 1.177 2.549= A y= 2.549 * .96^t 3
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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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q2. Find the exponential function corresponding to the points (5,3) and (10,2). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3=a*b^t/2=a*b^10 1.5=b^-5 b=.922 3=A*.922^5 a+4.503 y=4.503*.922^t confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qWhat are k1 and k2 such that b = e^k2 = 2^k1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: ********************************************* Question: `qIf one earthquake as an R value `dR higher than another, what is the ratio I2 / I1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: