course mth 163 ???????????ff???~?assignment #018018. `query 18
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19:09:00 query Logarithms, Logarithmic Functions, Logarithmic Equations 1. For what value of x will the function y = log{base 2}(x) first reach 4
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RESPONSE --> If we rewrite this : 4 = log base 2 (x) to find this: x = log 4 * log 2 x = (.301)(.602) x = .181 confidence assessment: 1
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19:20:33 ** 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. **
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RESPONSE --> I was still thinking about the last assignment with logs. (Since this happens to be from assignment 17 & and part of last week's assignement was from 18). I think Assignments 17 & 18 querys are messed up...may want to check on that. self critique assessment: 3
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19:22:31 for what value of x will the function y = ln(x) first reach y = 4?
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RESPONSE --> Is this not the same question? ln (x) = 4 x = e^4 x = approximately 55 confidence assessment: 1
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19:22:44 y = ln(x) means that e^y = x. The function y = ln(x) will first reach y = 4 when x = e^4 = 54.6 approx. **
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RESPONSE --> self critique assessment: 3
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19:35:32 3. 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
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RESPONSE --> I honestly can't find this in the notes, or by looking at that one graph. confidence assessment: 0
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19:41:35 ** 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. **
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RESPONSE --> After reading this explaination, it does make sense. I just didn't see where this information was found. self critique assessment: 1
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19:48:23 5. What are your estimates for the values of b for the two exponential functions on the given graph?
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RESPONSE --> My estimates would have to be maybe 1 and 2? confidence assessment: 0
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19:51:00 ** 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.. **
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RESPONSE --> I don't fully understand this one. I'll have to look back over this one. self critique assessment: 1
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20:09:46 7. What is the decibel level of a sound which is 10,000 times as loud as hearing threshold intensity?
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RESPONSE --> I am still not following these questions. As for the #7 question: If I read the notes correctly: log 10,000 = db 4 db confidence assessment: 1
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20:20:49 dB = 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.
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RESPONSE --> Ok, so I was on the right track. I didn't go quite to the right formula. dB = 10 log(I / I0). I also see where I/I0 = 10,000 and where it would be substituted in as: dB = 10 log(10,000) dB = 10 * 4 dB = 40 self critique assessment: 2
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20:23:50 What are the decibel levels of sounds which are 100, 10,000,000 and 1,000,000,000 times louder than threshold intensity?
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RESPONSE --> dB = 10 log(I / I0) dB = 10 log(100) dB = 10 *2 dB = 20 dB = 10 log(10,000,000) dB = 10 * 7 dB = 70 dB = 10 log(1,000,000,000) dB = 10 * 9 dB = 90 confidence assessment: 2
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20:23:57 10 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.
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RESPONSE --> self critique assessment: 3
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20:25:01 how can you easily find these decibel levels without using a calculator?
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RESPONSE --> count how many 0's are in the multiple of 10...that's the number (ex log 100 = 2) confidence assessment: 3
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20:25:09 Since 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.
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RESPONSE --> self critique assessment: 3
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20:30:37 What are the decibel levels of sounds which are 500, 30,000,000 and 7,000,000,000 times louder than threshold intensity?
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RESPONSE --> dB = 10 log(500) dB = 10 * 2.699 dB = 26.99 dB = 10 log(30,000,000) dB = 10 * 7.477 dB = 74.77 dB = 10 log(7,000,000,000) dB = 10 * 9.85 dB = 98.45 confidence assessment: 3
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20:30:43 10 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.
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RESPONSE --> self critique assessment: 3
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20:31:59 8. If a sound measures 40 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?
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RESPONSE --> dB = 10 log(I/I0) 40 = 10 log (x) 4 = log x x = 10000 confidence assessment: 2
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20:32:04 ** 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. **
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RESPONSE --> self critique assessment: 3
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20:34:42 Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.
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RESPONSE --> dB = 10 log(100) 20 = 10 log (I/I0) 2 = log (I/I0) 100 50 = 10 log (I/I0) 5 = log (I/I0) 100,000 80 = 10 log (I/I0) 8 = log (I/I0) 100,000,000 100 = 10 log (I/I0) 10 = log (I/I0) 10,000,000,000 confidence assessment: 2
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20:34:57 ** 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. **
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RESPONSE --> self critique assessment: 3
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20:36:07 What equation you would solve to find the intensity for decibel levels of 35, 83 and 117 dB.
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RESPONSE --> 35 = 10 log (I/I0) 83 = 10 log (I/I0) 117 = 10 log (I/I0) **Was not told that we needed to solve** confidence assessment: 2
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20:36:16 ** 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 **
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RESPONSE --> self critique assessment: 3
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20:47:24 9. is log(x^y) = x log(y) valid? If so why, and if not why not?
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RESPONSE --> no. If the equation were written log(x^y) = y log (x) then it would be fine. But the equation you gave is not valid confidence assessment: 1
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20:48:39 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **
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RESPONSE --> self critique assessment: 3
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20:51:26 is log(x/y) = log(x) - log(y) valid. If so why, and if not why not?
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RESPONSE --> yes it is valid. Accoring to the rules....it is possible. confidence assessment: 2
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20:52:00 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).
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RESPONSE --> self critique assessment: 3
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20:55:01 is log (x * y) = log(x) * log(y) valid. If so why, and if not why not?
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RESPONSE --> no, It would be log (x*y) = log x + log y confidence assessment: 2
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20:55:08 No. log(x * y) = log(x) + log(y)
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RESPONSE --> self critique assessment: 3
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20:57:28 03-26-2007 20:57:28 is 2 log(x) = log(2x) valid. If so why, and if not why not?
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NOTES -------> no, it would be 2 log (x) = log (x^2)
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20:57:30 is 2 log(x) = log(2x) valid. If so why, and if not why not?
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RESPONSE --> confidence assessment: 3
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20:57:37 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **
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RESPONSE --> self critique assessment: 3
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21:04:40 is log(x + y) = log(x) + log(y) valid. If so why, and if not why not?
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RESPONSE --> no, log (x+y) = log (ab) confidence assessment: 3
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21:04:49 ** log(x) + log(y) = log(xy), not log(x+y). **
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RESPONSE --> self critique assessment: 3
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21:05:50 is log(x) + log(y) = log(xy) valid. If so why, and if not why not?
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RESPONSE --> yeswe just proved it in the last question confidence assessment: 3
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21:06:08 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)
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RESPONSE --> self critique assessment: 3
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21:07:39 is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?
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RESPONSE --> no, log (x^y) = y log (x) confidence assessment: 3
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21:07:45 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)
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RESPONSE --> self critique assessment: 3
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21:09:23 is log(x - y) = log(x) - log(y) valid. If so why, and if not why not?
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RESPONSE --> no, log (x) - log (y) = log (a/b) confidence assessment: 3
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21:09:41 No. log(x-y) = log x/ log y
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RESPONSE --> self critique assessment: 3
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21:13:08 is 3 log(x) = log(x^3) valid. If so why, and if not why not?
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RESPONSE --> yes it is valid. confidence assessment: 1
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21:13:14 Yes. log(x^a) = a log(x).
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RESPONSE --> self critique assessment: 3
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21:14:53 is log(x^y) = y + log(x) valid. If so why, and if not why not?
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RESPONSE --> no, the correct answer would be log (x^y) = y log (x) confidence assessment: 3
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21:16:00 is log(x/y) = log(x) / log(y) valid. If so why, and if not why not?
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RESPONSE --> no, should be log (x/y) = log x - log y confidence assessment: 3
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21:16:14 is log(x^y) = y log(x) valid. If so why, and if not why not?
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RESPONSE --> yes confidence assessment: 3
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21:16:20 This is valid.
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RESPONSE --> self critique assessment: 3
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21:18:59 10. 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?
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RESPONSE --> This would be log {base 8} (1024) = log (1024) / log 8 (3.010) / (.903) 3.333 confidence assessment: 2
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21:21:33 COMMON 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.
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RESPONSE --> I don't fully understand where you got the log {base 8} (1024) = log {base 8) (2^10) and I really don't understand where you went after that. self critique assessment: 1
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21:34:31 what 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?
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RESPONSE --> log {base 2} (4 * 32) log {base 2} (4) + log {base 2} (32) log {base 2} (2^2) + log {base 2} (2^5) log {base 2} (2^2) = 2 log {base 2} (2^5) = 5 2 + 5 = 7 confidence assessment: 1
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21:34:44 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **
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RESPONSE --> self critique assessment: 3
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21:37:08 what do you get when you simplify log (1000)? If it can be evaluated exactly, what is the result and how did you get it?
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RESPONSE --> log (1000) = 3 A log without ""base"" is a base 10 log confidence assessment: 3
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21:38:28 what do you get when you simplify ln(3xy)? If it can be evaluated exactly, what is the result and how did you get it?
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RESPONSE --> If you can do ln like log: ln(3) +ln x + ln y confidence assessment: 1
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21:38:50 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)
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RESPONSE --> I just didn't simplify ln 3 self critique assessment: 2
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21:41:32 what 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?
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RESPONSE --> You get: .477 + .845 + 1.613 2.935 confidence assessment: 3
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21:44:39 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.
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RESPONSE --> I didn't realize that it had to be a power of 10. self critique assessment: 1
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21:53:34 11. 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.
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RESPONSE --> For log 12 (log 3 = .477) + 2 * (log 2^2 = .301) .477 + .301 + .301 1.079 There are no multiples of 17 especially any squares confidence assessment: 3
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22:18:18 12. What do you get when you solve 3 ^ (2x) = 7 ^ (x-4), and how did you solve the equation?
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RESPONSE --> 3 ^ (2x) = 7 ^ (x - 4) log [ 3 ^ (2x) ] = log [ 7 ^ (x - 4) ] (2x) log(3) = (x-4) log(7) 2x log(3) = x log(7) - 4 log(2) 2x log(3) - x log (7) = -4 log(7) x ( 2 log(3) - log(7) ) = -4 log(7) x = [-4 log(7)] / [ 2 log(3) - log(7)] x = (-3.38) / (.954 - .845) x = 3.38 / .109 x = 31.009 confidence assessment: 2
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22:20:02 ** 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. **
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RESPONSE --> I think I forgot the negative sign.
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22:34:32 What do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation?
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RESPONSE --> 2^(3x) + 2^(4x) = 9 log [ 2 ^ (3x) ] + log [ 2 ^ (4x) ] = 9 (3x) log(2) + (4x) log(2) = 9 3x log(2) + 4x log(2) = 9 3x log(2) + 4x log (2) = 9 x ( 3 log(2) +4 log(2) ) = 9 x = [9 / [ 3 log(2) + 4 log(2)] x = 9 / (.903 + 1.204) x = 9 / 2.107 x = 4.271 confidence assessment: 2
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22:35:53 COMMON 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. **
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RESPONSE --> so all that work for nothing. self critique assessment: 1
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22:40:13 What do you get when you solve 3^(2x-1) * 3^(3x+2) = 12, and how did you solve the equation?
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RESPONSE --> 3^(2x-1) * 3^(3x+2) = 12 3 ^ (2x-1+3x+2) = 12 3 ^ (5x + 1) = 12 confidence assessment:
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22:41:17 ** 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 **
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RESPONSE --> I accidentally hit the ""enter Response"" button. But I understand how to do this problem. self critique assessment: 3
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??????????????assignment #018 018. `query 18 Precalculus I 03-30-2007
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18:44:10 query 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?
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RESPONSE --> The exponential function I got was y = 2.587(2^-.053t) I got it by using the equations 3 = A (2^-4k1) and 2 = A(2^7k1) I divided the first by the second and got 1.5 = (2^-4k1)/(2^7k1) which then can be written: 1.5 = 2^(-4k1 - 7k1) or 1.5 = 2 ^ (-11k1) Then take the log: log 1.5 = -11k1 log 2 (log 1.5)/(-11 log 2) = k1 k1 = .176/-3.311 k1 = -.053 Then Substitute back into: 2 = A (2 ^ (7*-.053)) 2 = A (2 ^ (-.371) 2 = A (.773) A = 2.587 confidence assessment: 1
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18:47:53 ** 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). **
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RESPONSE --> I went back and rechecked my answer I still got that it was 2.587 (2^-.053t) self critique assessment: 1
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20:20:57 what is the exponential function of form A b^t such that the graph passes thru points thru points (-4,3) and (7,2).
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RESPONSE --> Somehow this keeps skipping on me. Here is the answer I was trying to type in: y = 2.591 e ^ (-.053t) First I used the equations: 3 = A * e^ (-4k2) and 2 = A * e ^ (7k2) Then divide the first by the second to get: 1.5 = (e ^ (-4k2)) / (e ^ 7k2) 1.5 = e ^ (-4k2 - 7k2) 1.5 = e ^ (-11k2) Take Natural Log: -11k2 ln e = ln 1.5 k2 = (ln 1.5) / (-11 ln e) k2 = .405 / -11 k2 = -.037 Put this back into an original equation: 2 = A * e ^(7 * -.037) 2 = A * e ^(-.259) 2 = A * .772 A = 2.591 I had a small problem with this one....but I am sure after looking at the explaination I will understand. I didn't get the equation, but here is how far I got: 3 = A*b^(-4) and 2 = A*b^(7) Divide 1st by the 2nd 1.5 = (b^(-4)) / (b^7) 1.5 = b ^(-4 - 7) 1.5 = b ^ (-11) *At this point I wasn't sure whether to make it linear or take the log to get b... confidence assessment: 1
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20:27:45 ** 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 **
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RESPONSE --> Ok, maybe not....I see the equation I started with but do not see how to get the .96. I know it is probably something simple that I should know, but I am having a memory lapse. self critique assessment: 1
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21:37:35 2. Find the exponential function corresponding to the points (5,3) and (10,2).
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RESPONSE --> I used the formula y = A * b^t 3 = A * b^5 and 2 = A *b^10 1.5 = (b^5) / (b^10) 1.5 = b ^ (5-10) 1.5 = b ^(-5) b= .922 3 = A * .922^5 3 = A *.666 A = 4.5 05 y = 4.505 * .922^t confidence assessment: 2
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21:37:54 ** 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= Ab^-5 b= .922 Now A = 3 / b^5 = 3 / .922^5 = 4.5. Our model is y = 4.5 * .922^t. **
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RESPONSE --> self critique assessment: 3
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22:05:11 What are k1 and k2 such that b = e^k2 = 2^k1?
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RESPONSE --> 3 = A * 2^(k1 * 5) and 2 = A * 2^(k1 * 10) 1.5 = (2^(k1*5) / (2^(k1*10) 1.5 = 2^ (5k1-10k1) 1.5 = 2^(-5k1) log 1.5 = -5k1 log 2 k1 = (log 1.5) / (-5 log 2) k1 = .176/-1.505 k1 = -.117 3 = A * 2^(-.117 * 5) 3 = A * 2^ (-.585) 3 = A * (.667) 4.5 = A y = 4.5 * 2^(.-117t) 3 = A * e^ (k2 *5) 2 = A * e^ (k2 *10) 1.5 = (e^(k2*5) / (e^(k2*10) 1.5 = e^ (5k2-10k2) 1.5 = 2^(-5k2) ln 1.5 = -5k2 ln e k2 = (ln 1.5) / (-5 ln e) k2 = ..405/-5 k2 = -.081 3 = A * 2^(-.081 * 5) 3 = A * 2^ (-.585) 3 = A * (.405) 7.407 = A y = 7.407 * e^(-.081t) b = .922 e^-.081 = .922 2^ -.117 = .992 confidence assessment: 3
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22:05:16 ** .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). ****
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RESPONSE --> self critique assessment: 3
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22:47:48 3. earthquakes measure R1 = 7.4 and R2 = 8.2. What is the ratio I2 / I1 of intensity and how did you find it?
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RESPONSE --> I wasn't sure how to find the I value from the equation: R = log (I/Io) I got this far: 7.4 = log (I/Io) and 8.2 = log (I/Io) The hind says solve for I....I'm not quite sure what I need to do. confidence assessment: 0
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22:57:59 ** 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. **
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RESPONSE --> After looking at the explaination, I totally understand how to do this! Great!!!! self critique assessment: 3
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23:09:37 ** 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. **
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RESPONSE --> Again it beat me to typing: Using the Info and what we know we can find: R2 = R1 + 1.6 and Using the formula, 10^ (R2-R1) 10^(R1+1.6-R1) = 10 ^ 1.6 Therefore it would be 39.81 times more intense. self critique assessment: 3
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23:10:00 If one earthquake as an R value `dR higher than another, what is the ratio I2 / I1?
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RESPONSE --> as with the other problem: 10^dR confidence assessment: 3
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23:10:06 ** As before I2 / I1 = 10^(R2-R1). If R2 is `dR greater than R1 we have R2 - R1 = `dR and I2 / I1 = 10^`dR. **
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RESPONSE --> self critique assessment: 3
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