course mth 163 Assignment 18 seemed to run into Assignment 19 and then go off somewhere else ... I stopped when I thought that it was getting into other assignments...let me know!
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13:42:32 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 --> x would be equal to 13.2877 4 = log{base 2} (x), I divided both sides by log {base 2} to come to x = 13.2877 then substituting it back into the original equation, I get 4 = log{base 2}(13.2877) 4 = .3010299957*13.2877 4 = 3.999999 or 4
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13:44:30 ** 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 --> Did I do this incorrectly...am I not understanding what you are asking me to complete...I follow what you are doing here, but my understanding of the question was that you were asking for the value of x in the equation 4 = log {base 2}(x). Please comment
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14:08:49 for what value of x will the function y = ln(x) first reach y = 4?
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RESPONSE --> y = ln(x) is 4 = ln(x) x = e^4 x = 54.598
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14:09:03 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 --> OK
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14:16:59 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 --> log{base b}(x) is the same the inverse to the function b^x. b raised to any power greater than 0 will be an asymptote for the x axis so the inverse of this graph will be an asymptote for the y axis. b has to be greater than 0...I don't understand why it could only be so if it was greater than 1...I can understand why it would be so if b less than or equal to 0 since you can't take the log of anything less than or equal to 0
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14:17:30 ** 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 --> OK
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14:18:51 5. What are your estimates for the values of b for the two exponential functions on the given graph?
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RESPONSE --> I don't have a graph to give an estimate for...I will continue to see if it is supplied
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14:22:49 ** 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 will have to ""pass"" on this one. I've read the answer, but since I had no graph to go by on this one, I'll just accept the answer and continue
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14:29:36 At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?
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RESPONSE --> I don't have a logarithmic function supplied for me to answer. I've copied down the request of the values of y and will continue to see if I can find the function that I will be able to solve
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15:27:27 ** 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. **
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RESPONSE --> using the problem y = 3.5^x i will solve for y = 2, 3, and 4. in the problem y = 3.5^x, in order for y = 2, x = .55 y = 3, x = .875 y = 4, x = 1.105 in the problem y = 7.3^x, in order for y = 2, x = .35 y = 3, x = .552 y = 4, x = .697 Basically the same figures you got, but in order to show that I did do the problem, I took it to the third significant number just to show.
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15:34:02 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 worked this problem, but the answer I get will not compute on the calculator, it tells me it is 0. If I'm correct, I'll explain, if incorrect, I'll self-critique.
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15:38:33 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 --> Well, I'm incorrect,,, This totally confused me when I first looked at it, but as I try to analyze what it is, I follow it, but still confused.
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15:43:04 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 --> if the decibel level of sounds is 100, the threshold intensity is 20. if the decibel level of sounds is 10,000,000, the threshold intensity is 70 if the decibel level of sounds is 1,000,000,000, the threshold intensity is 90
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15:43:17 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 --> OK
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15:47:14 how can you easily find these decibel levels without using a calculator?
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RESPONSE --> substitute the number is louder than (example: 100 times louder) I would put it in the form of y = 10^x, then multiply the power that 10 is raised to, and multiply by 10.
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15:47:26 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 --> OK
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15:50:15 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 --> threshold intensity^500 = Db is 26.989 or 27 threshold intensity^30,000,000 = Db is 74.77 threshold intensity^7,000,000,000 = Db is 98.45
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15:50:27 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 --> OK
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15:57:00 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 --> sound (40 Db) = 10*log{base thershold intensity}
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16:03:14 ** 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 --> well I guess that from my answer, I didn't understand the problem...here goes again... maybe I did understand the problem and just didn't simplify enough... 4 = log{base x} then translating (better word than I could think of) 10^4 = x...x would be 10,000
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16:06:48 Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.
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RESPONSE --> 20 decibels is 100 measures of sound 50 decibels is 100,000 measures of sound 80 decibels is 100,000,000 measures of sound 100 decibels is 10,000,000,000 measures of sound
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16:07:08 ** 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 --> OK
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16:11:00 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{base x} 3.5 = log{base x) 10^3.5 = x x = 3162.28 83 = 10*log{base x} 8.3 = log{base x} 10^8.3 = x x = 199526231.5 117 = 10*log{base x} 11.7 = log{base x} 10^11.7 = x
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16:11:23 ** 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 --> OK
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16:19:00 9. is log(x^y) = x log(y) valid? If so why, and if not why not?
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RESPONSE --> This is true according the laws of logarithms
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16:20:00 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **
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RESPONSE --> you are correct...I guess I need to review my laws of logarithms a little more closely.
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16:22:49 is log(x/y) = log(x) - log(y) valid. If so why, and if not why not?
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RESPONSE --> The law for log (x/y) is derived as log (x/y) = log (x y^-1) = log (x) + log (b^-1) = log (x) + (-1) log(y) = log (x) - log (y)
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16:23:00 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).
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RESPONSE --> OK
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16:24:34 is log (x * y) = log(x) * log(y) valid. If so why, and if not why not?
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RESPONSE --> No, this is not a valid statment...it should read log(x*y) = log (x) + log(y)
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16:24:41 No. log(x * y) = log(x) + log(y)
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RESPONSE --> OK
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16:30:08 is 2 log(x) = log(2x) valid. If so why, and if not why not?
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RESPONSE --> no, I really can't give you a specific reason, but I substitued several variables for x and the equations wouldn't equal.
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16:30:24 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **
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RESPONSE --> OK
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16:32:14 is log(x + y) = log(x) + log(y) valid. If so why, and if not why not?
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RESPONSE --> This is NOT a valid statement...the law of logarithms states that log(x) + log(y) = log(a*b)
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16:32:21 ** log(x) + log(y) = log(xy), not log(x+y). **
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RESPONSE --> OK
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16:32:49 is log(x) + log(y) = log(xy) valid. If so why, and if not why not?
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RESPONSE --> Yes, according to the law of logarithms
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16:33:07 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)
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RESPONSE --> OK
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16:34:46 is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?
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RESPONSE --> no, according to the law of logarithms log(x^y) = y log (x)
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16:34:55 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)
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RESPONSE --> OK
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16:36:08 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) is the inverse of log(x/y)
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16:36:19 No. log(x-y) = log x/ log y
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RESPONSE --> OK
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16:38:57 is 3 log(x) = log(x^3) valid. If so why, and if not why not?
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RESPONSE --> yes, according to the law of logarithms
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16:39:03 Yes. log(x^a) = a log(x).
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RESPONSE --> OK
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16:39:59 is log(x^y) = y + log(x) valid. If so why, and if not why not?
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RESPONSE --> no, log (x^y) = y log(x)
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16:40:05 No. log(x^y) = y log(x).
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RESPONSE --> OK
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16:41:01 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(x) - log(y)
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16:41:07 No. log(x/y) = log(x) - log(y).
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RESPONSE --> OK
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16:41:44 is log(x^y) = y log(x) valid. If so why, and if not why not?
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RESPONSE --> yes, according to the law of logarithms
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16:41:50 This is valid.
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RESPONSE --> OK
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16:49:03 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 --> log{base 8} (1024) = log 1024 / log 8 = 3.01 / .90 = 3.34
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16:56:20 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 guess it is a common error since you knew exacltly what I'd done prior to even seeing it...note taken.
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17:07:15 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 --> I'm hoping I did this one correctly...if my answer shows that I did so, I will explain, if not...it's self-critique time... my answer is 7
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17:17:20 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **
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RESPONSE --> YES!!!!! log{base 2}(4*32)= log{base 2} (2^7) = 7
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17:19:41 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 --> My answer is 3
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17:19:59 Since 10^3 = 1000, we have log (1000) = 3
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RESPONSE --> I just punched that one into the calculator
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17:22:07 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 --> I'll have to look to see...I'll self critique myself
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17:24:11 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)
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RESPONSE --> I was trying to make this harder than it is... In(3xy) is In(3) + In(x) + In(y)= 1.0986 + In(x) + In(y) As the blind man said, "" I see""
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17:28:10 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 --> My answer is 2.635
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17:28:47 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 see this, but I took the logs of each of these numbers and added them to come up with the answer provided.
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17:34:31 ** 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. **
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RESPONSE --> log 12 = log 10^1.079 which is 1.079
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21:31:20 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 --> ok...i'll have to look
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21:38:31 ** 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 --> my first mistake was not taking the log of both sides of the equations, once this is done all that is left is basic algebra to get to the point of finding the answer. I did this problem on paper twice, once watching each step and the second trying not to look at the first. I follow what happened.
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21:58:43 What do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation?
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RESPONSE --> `OK, I can't get this one either...I've tried it 4 times and 4 times I get the same answer, which I plug back into the original equation and it doesn't equal...I'm going to look
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22:01:15 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 --> maybe then I was not as dumb as I thought. I was making the common mistake that you listed.
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22:13:04 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) + (3x+2) = 12 multiply both sides by log(3) 2x-1 + 3x +2 = log{base 3} (12) 5x + 1 = 2.262 5x = 1.262 x = .2524
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22:13:27 ** 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 --> OK
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22:17:41 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 --> I'm totally confused...I will look further...
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22:28:22 ** 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 --> OK, I'm still confused...since I can't go back and I think I wrote the problem correctly...my problem read A(2^k1t) I follow what was done in the answer, but I'm having problems substituting in for the variables.
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22:36:00 what 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?
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RESPONSE --> trying again y = (.0275)(.898t)
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22:38:21 ** 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). **
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RESPONSE --> I was incorrect...big suprise... again...I'm lost
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22:45:15 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 --> y = 2.591 (.964)^t
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22:45:40 ** 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....""whew""
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22:54:29 2. Find the exponential function corresponding to the points (5,3) and (10,2).
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RESPONSE --> y = 1.524(1.145)^t
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22:58:12 ** 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 --> got the theory...just can't read... did it over (coping the problem correctly)...got the correct answer
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22:59:33 What are k1 and k2 such that b = e^k2 = 2^k1?
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RESPONSE --> I have absolutely no idea
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23:00:01 ** .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 --> OK
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"
course mth 163 Assignment 18 seemed to run into Assignment 19 and then go off somewhere else ... I stopped when I thought that it was getting into other assignments...let me know!
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13:42:32 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 --> x would be equal to 13.2877 4 = log{base 2} (x), I divided both sides by log {base 2} to come to x = 13.2877 then substituting it back into the original equation, I get 4 = log{base 2}(13.2877) 4 = .3010299957*13.2877 4 = 3.999999 or 4
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13:44:30 ** 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 --> Did I do this incorrectly...am I not understanding what you are asking me to complete...I follow what you are doing here, but my understanding of the question was that you were asking for the value of x in the equation 4 = log {base 2}(x). Please comment
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14:08:49 for what value of x will the function y = ln(x) first reach y = 4?
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RESPONSE --> y = ln(x) is 4 = ln(x) x = e^4 x = 54.598
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14:09:03 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 --> OK
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14:16:59 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 --> log{base b}(x) is the same the inverse to the function b^x. b raised to any power greater than 0 will be an asymptote for the x axis so the inverse of this graph will be an asymptote for the y axis. b has to be greater than 0...I don't understand why it could only be so if it was greater than 1...I can understand why it would be so if b less than or equal to 0 since you can't take the log of anything less than or equal to 0
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14:17:30 ** 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 --> OK
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14:18:51 5. What are your estimates for the values of b for the two exponential functions on the given graph?
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RESPONSE --> I don't have a graph to give an estimate for...I will continue to see if it is supplied
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14:22:49 ** 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 will have to ""pass"" on this one. I've read the answer, but since I had no graph to go by on this one, I'll just accept the answer and continue
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14:29:36 At what points will each of the logarithmic functions reach the values y = 2, y = 3 and y = 4?
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RESPONSE --> I don't have a logarithmic function supplied for me to answer. I've copied down the request of the values of y and will continue to see if I can find the function that I will be able to solve
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15:27:27 ** 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. **
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RESPONSE --> using the problem y = 3.5^x i will solve for y = 2, 3, and 4. in the problem y = 3.5^x, in order for y = 2, x = .55 y = 3, x = .875 y = 4, x = 1.105 in the problem y = 7.3^x, in order for y = 2, x = .35 y = 3, x = .552 y = 4, x = .697 Basically the same figures you got, but in order to show that I did do the problem, I took it to the third significant number just to show.
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15:34:02 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 worked this problem, but the answer I get will not compute on the calculator, it tells me it is 0. If I'm correct, I'll explain, if incorrect, I'll self-critique.
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15:38:33 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 --> Well, I'm incorrect,,, This totally confused me when I first looked at it, but as I try to analyze what it is, I follow it, but still confused.
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15:43:04 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 --> if the decibel level of sounds is 100, the threshold intensity is 20. if the decibel level of sounds is 10,000,000, the threshold intensity is 70 if the decibel level of sounds is 1,000,000,000, the threshold intensity is 90
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15:43:17 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 --> OK
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15:47:14 how can you easily find these decibel levels without using a calculator?
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RESPONSE --> substitute the number is louder than (example: 100 times louder) I would put it in the form of y = 10^x, then multiply the power that 10 is raised to, and multiply by 10.
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15:47:26 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 --> OK
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15:50:15 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 --> threshold intensity^500 = Db is 26.989 or 27 threshold intensity^30,000,000 = Db is 74.77 threshold intensity^7,000,000,000 = Db is 98.45
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15:50:27 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 --> OK
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15:57:00 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 --> sound (40 Db) = 10*log{base thershold intensity}
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16:03:14 ** 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 --> well I guess that from my answer, I didn't understand the problem...here goes again... maybe I did understand the problem and just didn't simplify enough... 4 = log{base x} then translating (better word than I could think of) 10^4 = x...x would be 10,000
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16:06:48 Answer the same question for sounds measuring 20, 50, 80 and 100 decibels.
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RESPONSE --> 20 decibels is 100 measures of sound 50 decibels is 100,000 measures of sound 80 decibels is 100,000,000 measures of sound 100 decibels is 10,000,000,000 measures of sound
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16:07:08 ** 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 --> OK
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16:11:00 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{base x} 3.5 = log{base x) 10^3.5 = x x = 3162.28 83 = 10*log{base x} 8.3 = log{base x} 10^8.3 = x x = 199526231.5 117 = 10*log{base x} 11.7 = log{base x} 10^11.7 = x
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16:11:23 ** 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 --> OK
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16:19:00 9. is log(x^y) = x log(y) valid? If so why, and if not why not?
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RESPONSE --> This is true according the laws of logarithms
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16:20:00 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **
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RESPONSE --> you are correct...I guess I need to review my laws of logarithms a little more closely.
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16:22:49 is log(x/y) = log(x) - log(y) valid. If so why, and if not why not?
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RESPONSE --> The law for log (x/y) is derived as log (x/y) = log (x y^-1) = log (x) + log (b^-1) = log (x) + (-1) log(y) = log (x) - log (y)
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16:23:00 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).
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RESPONSE --> OK
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16:24:34 is log (x * y) = log(x) * log(y) valid. If so why, and if not why not?
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RESPONSE --> No, this is not a valid statment...it should read log(x*y) = log (x) + log(y)
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16:24:41 No. log(x * y) = log(x) + log(y)
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RESPONSE --> OK
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16:30:08 is 2 log(x) = log(2x) valid. If so why, and if not why not?
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RESPONSE --> no, I really can't give you a specific reason, but I substitued several variables for x and the equations wouldn't equal.
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16:30:24 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **
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RESPONSE --> OK
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16:32:14 is log(x + y) = log(x) + log(y) valid. If so why, and if not why not?
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RESPONSE --> This is NOT a valid statement...the law of logarithms states that log(x) + log(y) = log(a*b)
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16:32:21 ** log(x) + log(y) = log(xy), not log(x+y). **
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RESPONSE --> OK
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16:32:49 is log(x) + log(y) = log(xy) valid. If so why, and if not why not?
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RESPONSE --> Yes, according to the law of logarithms
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16:33:07 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)
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RESPONSE --> OK
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16:34:46 is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not?
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RESPONSE --> no, according to the law of logarithms log(x^y) = y log (x)
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16:34:55 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)
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RESPONSE --> OK
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16:36:08 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) is the inverse of log(x/y)
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16:36:19 No. log(x-y) = log x/ log y
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RESPONSE --> OK
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16:38:57 is 3 log(x) = log(x^3) valid. If so why, and if not why not?
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RESPONSE --> yes, according to the law of logarithms
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16:39:03 Yes. log(x^a) = a log(x).
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RESPONSE --> OK
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16:39:59 is log(x^y) = y + log(x) valid. If so why, and if not why not?
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RESPONSE --> no, log (x^y) = y log(x)
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16:40:05 No. log(x^y) = y log(x).
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RESPONSE --> OK
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16:41:01 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(x) - log(y)
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16:41:07 No. log(x/y) = log(x) - log(y).
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RESPONSE --> OK
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16:41:44 is log(x^y) = y log(x) valid. If so why, and if not why not?
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RESPONSE --> yes, according to the law of logarithms
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16:41:50 This is valid.
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RESPONSE --> OK
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16:49:03 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 --> log{base 8} (1024) = log 1024 / log 8 = 3.01 / .90 = 3.34
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16:56:20 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 guess it is a common error since you knew exacltly what I'd done prior to even seeing it...note taken.
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17:07:15 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 --> I'm hoping I did this one correctly...if my answer shows that I did so, I will explain, if not...it's self-critique time... my answer is 7
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17:17:20 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **
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RESPONSE --> YES!!!!! log{base 2}(4*32)= log{base 2} (2^7) = 7
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17:19:41 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 --> My answer is 3
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17:19:59 Since 10^3 = 1000, we have log (1000) = 3
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RESPONSE --> I just punched that one into the calculator
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17:22:07 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 --> I'll have to look to see...I'll self critique myself
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17:24:11 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)
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RESPONSE --> I was trying to make this harder than it is... In(3xy) is In(3) + In(x) + In(y)= 1.0986 + In(x) + In(y) As the blind man said, "" I see""
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17:28:10 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 --> My answer is 2.635
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17:28:47 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 see this, but I took the logs of each of these numbers and added them to come up with the answer provided.
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17:34:31 ** 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. **
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RESPONSE --> log 12 = log 10^1.079 which is 1.079
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21:31:20 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 --> ok...i'll have to look
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21:38:31 ** 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 --> my first mistake was not taking the log of both sides of the equations, once this is done all that is left is basic algebra to get to the point of finding the answer. I did this problem on paper twice, once watching each step and the second trying not to look at the first. I follow what happened.
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21:58:43 What do you get when you solve 2^(3x) + 2^(4x) = 9, and how did you solve the equation?
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RESPONSE --> `OK, I can't get this one either...I've tried it 4 times and 4 times I get the same answer, which I plug back into the original equation and it doesn't equal...I'm going to look
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22:01:15 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 --> maybe then I was not as dumb as I thought. I was making the common mistake that you listed.
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22:13:04 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) + (3x+2) = 12 multiply both sides by log(3) 2x-1 + 3x +2 = log{base 3} (12) 5x + 1 = 2.262 5x = 1.262 x = .2524
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22:13:27 ** 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 --> OK
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22:17:41 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 --> I'm totally confused...I will look further...
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22:28:22 ** 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 --> OK, I'm still confused...since I can't go back and I think I wrote the problem correctly...my problem read A(2^k1t) I follow what was done in the answer, but I'm having problems substituting in for the variables.
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22:36:00 what 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?
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RESPONSE --> trying again y = (.0275)(.898t)
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22:38:21 ** 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). **
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RESPONSE --> I was incorrect...big suprise... again...I'm lost
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22:45:15 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 --> y = 2.591 (.964)^t
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22:45:40 ** 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....""whew""
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22:54:29 2. Find the exponential function corresponding to the points (5,3) and (10,2).
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RESPONSE --> y = 1.524(1.145)^t
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22:58:12 ** 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 --> got the theory...just can't read... did it over (coping the problem correctly)...got the correct answer
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22:59:33 What are k1 and k2 such that b = e^k2 = 2^k1?
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RESPONSE --> I have absolutely no idea
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23:00:01 ** .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 --> OK
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