course Mth 163 ڦz毳G}passignment #018
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01:05:04 ** 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 --> (x) = 4 translates to x = e^4, which occurs at x = 55 (x) = 2 translates to x = e^2, which occurs at x = 7.4 (x) = 3 translates to x = e^3, which occurs at x = 20 self critique assessment: 3
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01:05:35 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 --> y = ln(x) means that e^y = x. The function y = ln(x) will first reach y = 4 when x = e^4 = 54.6 self critique assessment: 3
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01:05:57 ** 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 --> 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 assessment: 3
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01:06:16 ** 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 --> 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 self critique assessment: 3
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01:06:34 ** 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 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 assessment: 3
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01:06:52 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 --> 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 self critique assessment: 3
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01:07:43 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 --> 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. self critique assessment: 3
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01:08:00 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 --> 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. self critique assessment: 3
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01:08:27 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 --> 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. self critique assessment: 3
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01:08:45 ** 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 --> 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 assessment: 3
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01:09:09 ** 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 --> 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 assessment: 3
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01:09:26 ** 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 --> 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 assessment: 3
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01:10:10 ** log(a^b) = b log a so log(x^y) should be y log (x), not x log(y). **
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RESPONSE --> log(a^b) = b log a so log(x^y) should be y log (x), not x log(y) self critique assessment: 3
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01:10:34 Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y).
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RESPONSE --> is log(x/y) = log(x) - log(y) valid. If so why, and if not why not? Yes , this is valid. It is the inverse of the exponential law a^x / a^y = a^(x-y). self critique assessment: 3
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01:10:49 No. log(x * y) = log(x) + log(y)
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RESPONSE --> is log (x * y) = log(x) * log(y) valid. If so why, and if not why not? No. log(x * y) = log(x) + log(y) self critique assessment: 3
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01:11:14 ** log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). **
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RESPONSE --> log(a^b) = b log a so 2 log(x) = log(x^2), not log(2x). self critique assessment: 3
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01:11:38 ** log(x) + log(y) = log(xy), not log(x+y). **
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RESPONSE --> log(x) + log(y) = log(xy), not log(x+y). self critique assessment: 3
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01:11:53 This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y)
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RESPONSE --> This is value. It is inverse to the law of exponents a^x*a^y = a^(x+y) self critique assessment: 3
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01:12:33 No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab)
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RESPONSE --> is log(x^y) = (log(x)) ^ y valid. If so why, and if not why not? No. log(x^y) = y log (x). This is the invers of the law (x^a)^b = x^(ab) self critique assessment: 2
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01:12:47 No. log(x-y) = log x/ log y
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RESPONSE --> is log(x - y) = log(x) - log(y) valid. If so why, and if not why not? No. log(x-y) = log x/ log y self critique assessment: 3
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01:13:02 Yes. log(x^a) = a log(x).
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RESPONSE --> is 3 log(x) = log(x^3) valid. If so why, and if not why not? Yes. log(x^a) = a log(x). self critique assessment: 3
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01:13:23 No. log(x^y) = y log(x).
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RESPONSE --> is log(x^y) = y + log(x) valid. If so why, and if not why not? No. log(x^y) = y log(x). self critique assessment: 3
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01:13:37 No. log(x/y) = log(x) - log(y).
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RESPONSE --> is log(x/y) = log(x) / log(y) valid. If so why, and if not why not? No. log(x/y) = log(x) - log(y). self critique assessment: 3
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01:13:49 This is valid.
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RESPONSE --> is log(x^y) = y log(x) valid. If so why, and if not why not? This is valid. self critique assessment: 3
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01:14:11 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 --> 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. self critique assessment: 3
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01:14:28 ** log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. **
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RESPONSE --> log{base 2}(4*32) = log{bse 2}(2^2 * 2^5) = log{base 2}(2^7) = 7. self critique assessment: 3
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01:15:24 Since 10^3 = 1000, we have log (1000) = 3
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RESPONSE --> Since 10^3 = 1000, we have log (1000) = 3 self critique assessment: 3
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01:15:50 ln(3xy) = ln (3) + ln(x) + ln(y) = 1.0986 + ln(x) + ln(y)
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RESPONSE --> ok self critique assessment: 3
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01:16:04 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 --> 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 assessment: 3
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01:16:26 ** 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 --> 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. self critique assessment: 3
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01:16:42 ** 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 --> 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 self critique assessment: 3
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01:17:12 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 --> log(a + b) is not equal to log(a) + log(b). log(a) + log(b) = log(a * b), not log(a + b). self critique assessment: 3
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01:17:49 ** 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 --> 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. x = .2524 self critique assessment: 3
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01:18:06 ** 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 --> 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 assessment: 3
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01:18:20 ** 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 --> 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 assessment: 3
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01:18:43 ** 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 --> 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 assessment: 3
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01:19:01 ** 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 --> 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. self critique assessment: 3
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01:19:18 ** .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 --> .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). self critique assessment: 3
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01:19:32 ** 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 --> 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) self critique assessment: 3
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01:19:46 ** 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 --> 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 assessment: 3
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01:19:59 ** 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 --> 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 assessment: 3
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