course Mth 163
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12:01:29 Linearizing Data and Curve Fitting Problem 1. table for y = 2 t^2 vs. t, for t = 0 to 3, linearize. Give your table and the table for sqrt(y) vs. t.
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RESPONSE --> The table for y vs. t would look as followed: t y 0 0 1 2 2 8 3 18 The table for sqrt(y) vs t would look like: t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 confidence assessment: 2
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12:11:57 ** The table for y vs. t is t y 0 0 1 2 2 8 3 18 The table for sqrtIy) vs t, with sqrt(y) give to 2 significant figures, is t sqrt(y) 0 0 1 1.4 2 2.8 3 4.2 **
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RESPONSE --> This was a fairly easy question. I got it correct. self critique assessment: 3
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12:22:25 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> The sqrt(y) sequence would be 0, 1.4, 2.8, 4.2 Therefore, the first-difference sequence would be 1.4, 1.4, 1.4, showing it is constant and nonzero. confidence assessment: 3
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12:38:47 The sqrt(y) sequence is 0, 1.4, 2.8, 4.2. The first-difference sequence is 1.4, 1.4, 1.4, which is constant and nonzero.
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RESPONSE --> Answered it correctly. confidence assessment: 3
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12:50:41 Give your values of m and b for the linear function that models your table.
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RESPONSE --> The points (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points show a straight line through the origin with a slope of 1.4, so the equation of the line is sqrt(y) = 1.4 t + 0 or simply just sqrt(y)=1.4 confidence assessment: 2
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12:59:40 ** The points (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points are fit by a straight line thru the origin with slope 1.4, so the equation of the line is sqrty) = 1.4 t + 0, or just sqrt(y) = 1.4 t. **
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RESPONSE --> This one was a bit more confusing, but I got it. However, I left the t off of the final solution and I need to pay attention to little things like that more because it could cause me to miss a question. self critique assessment: 3
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13:08:38 Does the square of this linear functiongive you back the original function?
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RESPONSE --> In squaring both sides of sqrt(y) = 1.4 t I got y = 1.96 t^2 The original function was y = 2 t^2. 1.96 would round off to two, therefore, yes the function does give back the original function as far as I can tell. confidence assessment: 3
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13:15:05 ** Squaring both sides of sqrt(y) = 1.4 t we get y = 1.96 t^2. The original function was y = 2 t^2. Our values for the sqrt(y) function were accurate to only 2 significant figures. To 2 significant figures 1.96 would round off to 2, so the two functions are identical to 2 significant figures. *&*&
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RESPONSE --> Wow. Yes, i got it right. I explained it in a confusing way, but I still got ir right. self critique assessment: 3
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13:21:31 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> confidence assessment:
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13:30:41 ** A table for the function is t y = 7 ( 3^t) 0 7 1 21 2 63 3 189 The table for log(y) vs. t is t log(7 ( 3^t)) 0 0..85 1 1.32 2 1.80 3 2.28/ Sequence analysis on the log(7 * 3^t) values: sequence 0.85 1.32 1.80 2.28 1st diff .47 .48 .48 The first difference appears constant with value about .473. log(y) is a linear function of t with slope .473 and vertical intercept .85. We therefore have log(y) = .473 t + .85. Thus 10^(log y) = 10^(.473 t + .85) so that y = 10^(.473 t) * 10^(.85) or y = (10^.473)^t * (10^.85), which evaluating the power of 10 with calculator gives us y = 2.97^t * 7.08. To 2 significant figures this is the same as the original function y = 3 * 7^t. **
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RESPONSE --> self critique assessment:
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13:38:18 problem 7. Hypothesized fit is `sqrt(y) = 2.27 x + .05. Compare your result to the 'ideal' y = 5 t^2 function.
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RESPONSE --> The y values are: .14, 5.66, 23.2, 52, 82.2 and 135.5. The square roots of these values would be: 0.374; 2.38, 4.82; 7.21; 9.34, 11.64. When plotting these square roots vs. t = 0, 1, 2, 3, 4, 5, I got a graph that resembles almost a straight line. confidence assessment:
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13:47:05 ** For the simulated data the y values are .14, 5.66, 23.2, 52, 82.2 and 135.5. The square roots of these values are 0.374; 2.38, 4.82; 7.21; 9.34, 11.64. Plotting these square roots vs. t = 0, 1, 2, 3, 4, 5 we obtain a nearly straight-line graph. The best-fit linear function to sqrt(y) vs. x gives us sqrt(y) = 2.27t + 0.27. Your function should be reasonably close to this but will probably not be identical. Squaring both sides we get y = 5.1529t^2 + 1.2258t + 0.0729. If the small term .0729 is neglected we get y = 5.15 t^2 + 1.23 t. Because of the 1.23 t term this isn't a particularly good approximation to y = 5 t^2.**
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RESPONSE --> I accidently submitted before I finished putting in my solution. I do understand everything about the problem though because I actually wasn't positive about how to do it. self critique assessment: 0
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13:56:59 problem 9. Assuming exponential follow the entire 7-step procedure for given data set Give your x and y data. Show you solution. Be sure to give the average deviation of your function from the given data?
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RESPONSE --> For (t, y) data set: (0,.42),(1,.29),(2,.21),(3,.15),(4,.10),(5,.07),I get log(y) vs. t: Therefore it would look like this: t log(y) 0 -.375 1 -.538 2 -.678 3 -.824 4 -1 5 -1.15 Therefore, log(y) = - 0.155x - 0.374 When solving I got 10^log(y) = 10^(- 0.155t - 0.374) or y = 10^-.374 * (10^-.155)^t or y = .42 * .70^t The columns below give t, y as in the original table, y calculated as y = .42 * .70^t and the difference between the predicted and original values of y: 0 0.42 0.42 0 1 0.29 0.294 -0.004 2 0.21 0.2058 0.0042 3 0.15 0.14406 0.00594 4 0.1 0.100842 -0.000842 5 0.07 0.0705894 -0.0005894 Deviations in the last column have an average value of -.00078 confidence assessment: 2
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14:18:13 For (t, y) data set (0,.42), (1,.29),(2,.21),(3,.15),(4,.10),(5,.07) we get log(y) vs. t: t log(y) 0 -.375 1 -.538 2 -.678 3 -.824 4 -1 5 -1.15 A best fit to this data gives log(y) = - 0.155x - 0.374. Solving we get 10^log(y) = 10^(- 0.155t - 0.374) or y = 10^-.374 * (10^-.155)^t or y = .42 * .70^t. The columns below give t, y as in the original table, y calculated as y = .42 * .70^t and the difference between the predicted and original values of y: 0 0.42 0.42 0 1 0.29 0.294 -0.004 2 0.21 0.2058 0.0042 3 0.15 0.14406 0.00594 4 0.1 0.100842 -0.000842 5 0.07 0.0705894 -0.0005894 The deviations in the last column have an average value of -.00078. This indicates that the model is very good. **
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RESPONSE --> THumbs Up! self critique assessment: 3
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14:25:00 problem 11. determine whether the log(y) vs. t or the log(y) vs. log(t) transformation works. Complete the problem and give the average discrepancy between the first function and your data.
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RESPONSE --> confidence assessment:
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14:36:04 ** The first table gives us x y log(x) log(y) 0.5 0.7 -0.30103 -0.1549 1 0.97 0 -0.01323 1.5 1.21 0.176091 0.082785 2 1.43 0.30103 0.155336 2.5 1.56 0.39794 0.193125 log(y) vs. x is not linear. log(y) vs. log(x) is linear with equation log(y) = 0.5074 log(x) - 0.0056. Applying the inverse transformation we get 10^log(y) =10^( 0.5074 log(x) - 0.0056) which we simplify to obtain y = 0.987x^0.507. The second table gives us x y log(x) log(y) 2 2.3 0.30103 0.361728 4 5 0.60206 0.69897 6 11.5 0.778151 1.060698 8 25 0.90309 1.39794 log(y) vs. x is linear, log(y) vs. log(x) is not. From the linear graph we get log(y) = 0.1735x + 0.0122, which we solve for y: 10^log(y) = 10^(0.1735x + 0.0122) or y = 10^.0122 * 10^(0.1735x) = 1.0285 * 1.491^x. **
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RESPONSE --> self critique assessment:
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14:43:30 Inverse Functions and Logarithms, Problem 7. Construct table for the squaring function f(x) = x^2, using x values between 0 and 2 with a step of .5. Reverse the columns of this table to form a partial table for the inverse function.
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RESPONSE --> The table is as follows: x f(x) 0 0 .5 .25 1 1 1.5 2.25 2 4 After reversing columns I got the following partial table for the inverse function: x f^-1(x) 0 0 .25 .5 1 1 2.25 1.5 4 2 confidence assessment: 2
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14:55:39 ** The table is x f(x) 0 0 .5 .25 1 1 1.5 2.25 2 4. Reversing columns we get the following partial table for the inverse function: x f^-1(x) 0 0 .25 .5 1 1 2.25 1.5 4 2 **
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RESPONSE --> self critique assessment: 3
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15:06:04 Describe your graph consisting of the smooth curves corresponding to both functions. How are the pairs of points positioned with respect to the y = x function?
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RESPONSE --> The curve of the original function is increasing at an increasing rate, the curve for the inverse function is increasing at a decreasing rate. In conclusion, the curves meet at (0, 0) and at (1, 1) The line connecting the pairs of points passes through the y = x line at a right angle, and the y = x line bisects each connecting line; therefore, the two graphs are mirror images of one another with respect to the line y = x confidence assessment: 3
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15:11:11 ** The curve of the original function is increasing at an increasing rate, the curve for the inverse function is increasing at a decreasing rate. The curves meet at (0, 0) and at (1, 1). The line connecting the pairs of points passes through the y = x line at a right angle, and the y = x line bisects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x. **
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RESPONSE --> self critique assessment: 3
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15:21:49 8. If we reversed the columns of the 'complete' table of the squaring function from 0 to 12, precisely what table would we get?
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RESPONSE --> A reversed table would give me the table for the square root function y = sqrt(x) The y = x^2 and y = sqrt(x) functions are inverse functions for x >= 0 confidence assessment: 3
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15:26:55 ** Our reversed table would give us the table for the square root function y = sqrt(x). The y = x^2 and y = sqrt(x) functions are inverse functions for x >= 0. **
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RESPONSE --> self critique assessment: 3
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15:29:21 9. If we could construct the 'complete' table of the squaring function from 0 to infinity, listing all possible positive numbers in the x column, then why would we be certain that every possible positive number would appear exactly one time in the second column?
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RESPONSE --> The second column consists of all the squares. In order for a number to appear in the second column, the square root of that number would have to appear in the first column. every possible number appears in the first column, so no matter what number is selected, it will appear in the second column. Therefore, every possible number appears in the second column. However, no number can appear more than once in the first column. confidence assessment: 2
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15:35:44 ** The second column consists of all the squares. In order for a number to appear in the second column the square root of that number would have to appear in the first. Since every possible number appears in the first column, then no matter what number we select it will appear in the second column. So every possible positive number appears in the second column. If a number appears twice in the second column then its square root would appear twice in the first column. But no number can appear more than once in the first column. **
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RESPONSE --> I left some things out, but overall I understood the explanation. self critique assessment: 3
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15:45:27 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> Next to 4.31 in the first column, 4.31^2 = 18.5761 would appear in the second column. confidence assessment: 3
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15:50:32 ** The table is the squaring function so next to 4.31 in the first column, 4.31^2 = 18.5761 would appear in the second. **
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RESPONSE --> self critique assessment: 3
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15:55:03 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> The square of sqrt(18) is 18, therefore, 18 would appear in the second column. confidence assessment: 3
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15:59:08 ** The square of sqrt(18) is 18, so 18 would appear in the second column. **
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RESPONSE --> ok self critique assessment: 3
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16:10:23 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> pi^2 would appear in the second column confidence assessment: 3
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16:19:27 ** pi^2 would appear in the second column. **
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RESPONSE --> self critique assessment: 3
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16:30:25 What would we obtain if we reversed the columns of this table?
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RESPONSE --> The table would have the square of the second column value in the first column, hence, the second column would be the square root of the first column! The function would now be the square-root function. confidence assessment: 3
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16:35:33 Our table would have the square of the second-column value in the first column, so the second column would be the square root of the first column. Our function would now be the square-root function.
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RESPONSE --> Got it! self critique assessment: 3
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16:39:58 What number would appear in the second column next to the number 4.31 in the first column of this table?
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RESPONSE --> sqrt(4.31) = 2.076 confidence assessment: 3
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16:42:03 ** you would have sqrt(4.31) = 2.076 **
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RESPONSE --> self critique assessment: 3
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16:50:32 What number would appear in the second column next to the number `pi^2 in the first column of this table?
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RESPONSE --> The number in the second column would be pi confidence assessment: 3
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16:56:40 ** The number in the second column would be pi, since the first-column value is the square of the second-column value. **
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RESPONSE --> I got it! self critique assessment: 3
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17:15:42 What number would appear in the second column next to the number -3 in the first column of this table?
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RESPONSE --> I don't understand this question. confidence assessment: 0
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17:33:13 ** -3 would not appear in the first column of the reversed table of the squaring function, since it wouldn't appear in the second column of that table. **
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RESPONSE --> OH OKAY...I see. It was sort of like a trick question. self critique assessment: 3
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17:44:52 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18
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RESPONSE --> confidence assessment:
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17:49:57 ** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base 2}(18) = log(18) / log(2). **
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RESPONSE --> self critique assessment:
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17:54:15 2 ^ (4x) = 12
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RESPONSE --> confidence assessment:
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18:09:20 ** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as 4 x = log{base 2}(12) = log(12) / log(2). **
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RESPONSE --> self critique assessment:
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18:18:48 5 * 2^x = 52
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RESPONSE --> I got 2^x = 52/5 therefore, x = log{base 2}(52/5) = log(52/5) / log(2) confidence assessment: 3
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18:25:56 ** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **
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RESPONSE --> self critique assessment: 3
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18:25:30 2^(3x - 4) = 9.
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RESPONSE --> I got: 3x - 4 = log 9 / log 2 therefore, 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3 confidence assessment: 3
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18:35:33 ** You get 3x - 4 = log 9 / log 2 so that 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3. **
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RESPONSE --> self critique assessment: 3
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18:48:05 14. Solve each of the following equations: 2^(3x-5) + 4 = 0
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RESPONSE --> I got log(-4)/log(2)=3x - 5 confidence assessment: 3
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18:55:24 ** You get log(-4)/log(2)=3x - 5. However log(-4) is not a real number so there is no solution. Note that 2^(3x-5) cannot be negative so the equation is impossible. **
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RESPONSE --> Ok okay. I get this now. self critique assessment: 3
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18:59:01 2^(1/x) - 3 = 0
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RESPONSE --> I got 2^(1/x) = 3 therefore 1/x = log(3) / log(2) and x = log(2) / log(3) = .63 confidence assessment: 3
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19:07:08 ** You get 2^(1/x) = 3 so that 1/x = log(3) / log(2) and x = log(2) / log(3) = .63 approx. **
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RESPONSE --> CORRECT self critique assessment: 3
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19:15:56 2^x * 2^(1/x) = 15
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RESPONSE --> 2^x * 2^(1/x)=x + 1/x = log{base 2}(15) Then multiplying both sides by x I got =x^2 + 1 = log{base 2}(15) Must be rearranged to get x^2 - log{base 2}(15) x + 1 = 0 Solutions: x = 0.2753664762 or x = 3.631524119 confidence assessment: 2
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19:26:00 ** 2^x * 2^(1/x) is the same as 2^(x + 1/x) so you get x + 1/x = log{base 2}(15). Multiplying both sides by x we get x^2 + 1 = log{base 2}(15). This is quadratic. We rearrange to get x^2 - log{base 2}(15) x + 1 = 0 then use quadratic formula with a=1, b=-log{base 2}(15) and c=4. Our solutions are x = 0.2753664762 OR x = 3.631524119. **
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RESPONSE --> self critique assessment: 3
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19:39:30 (2^x)^4 = 5
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RESPONSE --> I take the 1/4 power of both sides to get 2^x = 5^(1/4) therefore, x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2) confidence assessment: 3
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19:44:40 ** You take the 1/4 power of both sides to get 2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2). **
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RESPONSE --> self critique assessment: 3
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