course mth 163 I just wanted to let you know that I have submitted my test in the learing lab i took it last Thursday and was wondering when grades will be posted? Ñ»’š ÆÙS‚ñ‚îÕ «‰´Íß“âa‹assignment #017
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16:25:31 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 --> x column =0,1,2,3 and the y column=0,2,8,18 and the sqrt(y)=(0,0) (1,1.41) (2,2.8) (3,4.24) confidence assessment: 3
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16:25:40 ** 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 --> ok self critique assessment:
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16:26:05 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> yes because they all increase by 1.4 confidence assessment: 3
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16:26:12 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 --> ok confidence assessment: 3
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16:26:29 Give your values of m and b for the linear function that models your table.
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RESPONSE --> y=1.4x(t) confidence assessment: 3
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16:27:13 ** 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 --> I didnt do the sqrt in front of it but i understand why and how we got the 1.4 and the slope connected self critique assessment: 3
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16:27:39 Does the square of this linear functiongive you back the original function?
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RESPONSE --> No not if you square both sides confidence assessment: 3
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16:28:14 ** 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 --> oh ok i didnt test it but i didnt think logically that it would work self critique assessment: 3
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16:31:01 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> I dont know quite how to linearize this set but i did find a common change that they are all multiplied by 3 confidence assessment: 3
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16:32:59 ** 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 --> Ok this is starting to make sense you can go from any equation and take the reverse of it and ""linearize"" it I actually never thought about using the log on this one because i was thrown off by it just being raised to x self critique assessment: 3
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16:36:31 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 --> This was close but the actual equation i got was sqrt(y)=2.236x and it squares back out to equal 4.999 which is very close to the 5 confidence assessment: 3
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16:38:28 ** 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) = 2.27·t + 0.27. Your function should be reasonably close to this but will probably not be identical. Squaring both sides we get y = 5.1529·t^2 + 1.2258·t + 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 --> ok i dont really understand where the 1.23 came from in the first place self critique assessment: 3
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16:39:23 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 --> Im completely lost on this one confidence assessment: 2
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16:41:31 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) = - 0.155·x - 0.374. Solving we get 10^log(y) = 10^(- 0.155·t - 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 --> I was really lost on this concept on the worksheet but seeing it like this actually laid out in front of me really helped... self critique assessment: 3
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16:44:14 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: 3
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16:45:13 ** 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 = 0.987·x^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^(0.1735·x) = 1.0285 * 1.491^x. **
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RESPONSE --> I am lost on where these tables and data are coming from I did most of the worksheets prior to this and I am not understanding where all these questions can be found within them self critique assessment: 3
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16:46:49 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 --> (0,0) (1,1) (2,4) and then (0,0) (1,1) (4,2) confidence assessment: ok
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16:47:40 ** 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 --> ok i understand the .5 step now i missed the .5 step part but I did do these the correct way self critique assessment: 3
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16:49:56 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 points are the same until you get to certain points on the negative and postive graph and then as they approach higher positive numbers the x^2 function curves upward and the reversed function curves downward confidence assessment: 3
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16:50:03 ** 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 --> ok self critique assessment: 3
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16:54:40 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 --> x y 0 0 1 1 4 2 9 3 16 4 25 5 36 6 49 7 64 8 81 9 100 10 121 11 144 12 confidence assessment: 3
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16:54:52 ** 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 --> ok self critique assessment: 3
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16:55:50 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 --> because we would be going to infinity and not going backwards there would be no negatives and the numbers would continuously get higher confidence assessment: 3
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16:56:03 ** 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 --> ok self critique assessment: 3
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16:56:33 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> 2.076 confidence assessment: 3
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16:57:03 ** 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 --> i thought that 4.31 had been squared and was in the reverse row 1 column self critique assessment: 3
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16:57:20 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> 4.24 confidence assessment: 3
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16:57:49 ** The square of sqrt(18) is 18, so 18 would appear in the second column. **
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RESPONSE --> wow this is really kindof tricky i need to start really thinking before i type self critique assessment: 3
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16:58:23 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> 9.85 confidence assessment: 3
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16:58:42 ** pi^2 would appear in the second column. **
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RESPONSE --> ok i literally took 3.14 and squared it self critique assessment: 3
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16:59:18 What would we obtain if we reversed the columns of this table?
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RESPONSE --> The same one as before it was just covering more numbers confidence assessment: 3
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16:59:29 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 --> ok self critique assessment: 3
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16:59:41 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 --> 18 confidence assessment: 3
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16:59:51 ** you would have sqrt(4.31) = 2.076 **
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RESPONSE --> self critique assessment: 3
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17:00:03 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 --> pi confidence assessment: 3
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17:00:11 ** 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 --> ok finally self critique assessment: 3
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17:00:54 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 --> 1.73i confidence assessment: 3
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17:01:11 ** -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 wow that was really tricky self critique assessment: 3
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17:01:38 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18
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RESPONSE --> 4.1699 confidence assessment: 3
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17:01:56 ** 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 --> this is how i solved it self critique assessment: 3
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17:02:42 2 ^ (4x) = 12
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RESPONSE --> log(12)/log(2) and then divide by 4 and you get 0.0896 confidence assessment: 3
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17:02:49 ** 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 --> ok self critique assessment: 3
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17:04:17 5 * 2^x = 52
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RESPONSE --> i actually forgot how to do this one confidence assessment: 3
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17:04:28 ** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **
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RESPONSE --> ok self critique assessment: 3
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17:05:00 2^(3x - 4) = 9.
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RESPONSE --> log(9)/log(2)=3x-4 confidence assessment: 3
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17:05:11 ** 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 --> ok self critique assessment: 3
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17:05:52 14. Solve each of the following equations: 2^(3x-5) + 4 = 0
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RESPONSE --> 2.333 confidence assessment: 3
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17:06:30 ** 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 --> oh ok i missed the negative self critique assessment: 3
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17:07:28 2^(1/x) - 3 = 0
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RESPONSE --> .632 confidence assessment: 3
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17:07:34 ** 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 --> ok self critique assessment: 3
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17:07:58 03-27-2009 17:07:58 2^x * 2^(1/x) = 15
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NOTES -------> i have no clue on this one
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17:07:58 2^x * 2^(1/x) = 15
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RESPONSE --> confidence assessment:
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17:08:27 ** 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 --> wow ok self critique assessment: 3
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17:09:20 (2^x)^4 = 5
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RESPONSE --> 7.288 confidence assessment: 3
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17:10:39 ** 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 --> i took the sqrt of 4 versus taking a fourth of it and i kindof understand that but not completely self critique assessment: 3
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