course Mth 163
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19:48:53 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 --> T Y -2 8 -1 2 0 0 1 2 2 8 3 18
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19:51:23 ** 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 --> I forgot to do both tables...I will construct the sqrt(y) table before I scroll down to look T Y -2 2.83 -1 1.41 0 0 1 1.41 2 2.83 3 4.24 I will scroll now... I am correct
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19:53:42 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> yes
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19:54:31 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 --> I was a little more specific in my calculations...I went out to 3 significant numbers, but it corresponds to my answer.
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20:04:19 Give your values of m and b for the linear function that models your table.
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RESPONSE --> The linear equation is y = mx + b 4.24 = 3m + 1.4 I really don't know, this is my first thought, but I will continue and see where I'm at.
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20:09:29 ** 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 --> How simple...this makes a whole lot more sense than what I was trying to do, and if I could go back and change it I would, but... when I plotted my points, I found that the straight line does go thru them so the equation would be y(in this case sqrty) = 1.4t + 0
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20:13:31 Does the square of this linear functiongive you back the original function?
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RESPONSE --> no
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20:24:14 ** 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 --> I was doing my ""checking"" incorrectly...I was squaring 2t^2, therefore my answers were not identical...after seeing the answer, I see that they would be identical, or close enough to call identical.
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20:34:18 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> I'll have to say that I can't linearize this function. I can't get a get the same slope between the points even when I use the sqrt(y). y = 7(3^t) t y -2 .78 -1 2.33 0 7 1 21 2 63 t sqrt(y) -2 .88 -1 1.53 0 2.65 1 4.58 2 7.94
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20:44:52 ** 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 * (10^.85)^t, which evaluating the power of 10 with calculator gives us y = 2.97 * 7.08^t. To 2 significant figures this is the same as the original function y = 3 * 7^t. **
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RESPONSE --> WOW!!!! That was confusing...I'm going to take a minute and go over this before commenting... looking at your answer, I was able to come up with the same table and therefore the same answer...I am a little confused on when to use ""log"" and when to use ""sqrt(y)...I will assume unless you tell me different, that if one doesn't work...use the other???
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21:38:55 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 hypothesized fit of sqrt(y) = 2.27 x + .05 is the same as the ""ideal"" function of y = 5 t^2...once both sides were squared the function came out the same...the vertical intercept of ..0025 was so close to 0 that I used ""0"" .
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21:39:19 ** 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 --> OK
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21:40:07 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 --> I don't see a data set in this problem ????
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22:37:15 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 --> Although I had to look at the answer to get the problem, I did the problem myself and came up with a slightly different answer than the one listed: y = .42 * .70^t t y 0 .42 1 .294 2 .2058 3 .14406 4 .100842 5 .0705894 t log(y) 0 -.37675 1 -.53165 2 -.68655 3 -.84146 4 -.99636 5 -1.15126 The difference between the log(y) being -.1549 gives me the equation log(y) = .1549x - .37675 solving: 10^log(y) = 10^(-.1549x - .3768) y = 10^ -.3768 * 10^ -.1549^t y = .41976 * .6998^t y = .42 * .70^t
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22:38:01 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 --> I see no problem, therefore I will have to look further...
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22:40:47 ** 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 = 0.994x^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 = 1.012274723^(0.1735x). **
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RESPONSE --> I read this but simply don't know what you're asking me to do ???
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22:44:31 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 --> x y 0 0 .5 .25 1 1 1.5 2.25 2 4 Inverse table x y 0 0 .25 .5 1 1 2.25 1.5 4 2
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22:44:44 ** 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
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22:50:30 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 --> Using the points listed on the previous question, I plotted a graph with these points. Using a straight line y = x running through the middle of them, there is the same amount of distance between the smooth curves of the listed points and the straight line on either side of the staight line. If a line drawn from the smooth curved graph thru the straight line, they would intersect the straight line at a a 90 degree angle.
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22:50:45 ** 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
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22:54:28 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 --> If I'm understanding the question properly, we would get the sqrt(y) table.
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22:54:39 ** 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
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22:56:00 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 --> ??? There is only one sqrt(x) for every possible positive number > 0
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22:56:21 ** 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
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22:57:08 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> 4.31 squared is 18.5761
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22:57:16 ** 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 --> OK
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22:58:28 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> 18^ 2 is 324
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22:59:24 ** The square of sqrt(18) is 18, so 18 would appear in the second column. **
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RESPONSE --> of course...you would square the number in the 1st column, so the sqrt(x) would always be x
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23:00:16 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> 329791.8304
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23:00:47 ** pi^2 would appear in the second column. **
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RESPONSE --> you are correct...
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23:02:03 What would we obtain if we reversed the columns of this table?
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RESPONSE --> I'm not sure what table that you are referring to?
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23:03:13 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 --> oh...of the original f(x) = x^2...if these columns were reversed you would get the sqrt(x)
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23:04:41 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 --> table y = x^2 : 18.5761
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23:05:10 ** you would have sqrt(4.31) = 2.076 **
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RESPONSE --> sqrt (4.31) = 2.0761
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23:05:38 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
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23:05:45 ** 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
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23:07:19 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 --> you can't take the sqrt of a negative number, therefore it is undefined
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23:07:30 ** -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. ** 22:32:47
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RESPONSE --> OK
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23:08:31 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible:
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RESPONSE --> I don't have any equations to solve here
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23:13:35 2 ^ x = 18
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RESPONSE --> If I'm correct, I'll explain ... x = 4.169
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23:14:13 ** 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 --> exactly as I did, my answer being 17.9885 rounded off to 18
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23:18:16 2 ^ (4x) = 12
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RESPONSE --> 4x = log 12 / log 2 4x = 3.5850 x = .89626
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23:18:23 ** 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
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23:22:15 5 * 2^x = 52
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RESPONSE --> x = 3.3787 x = log 10.4 / log 2
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23:22:27 ** 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
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23:31:13 2^(3x - 4) = 9.
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RESPONSE --> x = 2.38998 3x-4 = log 9 / log 2
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23:31:21 ** 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
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23:34:26 14. Solve each of the following equations: 2^(3x-5) + 4 = 0
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RESPONSE --> I'll have to look at the answer to get this one...
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23:35:40 ** 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 --> just what I thought...I tried the log of -4 and it was an unreal number, therefore I thought maybe I was trying to do it incorrectly...I had to look and I was correct.
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23:39:46 2^(1/x) - 3 = 0
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RESPONSE --> x = .63093
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23:39:56 ** 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
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23:47:01 2^x * 2^(1/x) = 15
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RESPONSE --> I know it is not correct...but it is close...x = 3.9
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20:25:35 ** 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 --> I did this one wrong...I got to this page to try to follow your work and I can't ... I must have written the problem incorrectly on my paper...I don't get a quadratic equation... following your example I still have problems ???
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20:33:28 (2^x)^4 = 5
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RESPONSE --> 4x = log5/log 2 x = .5805
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20:37:01 ** 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 got the same answer, although I did it another way...whether this is a coinsidence or practical, I don't know.
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