course mth 163 I didn't understand why the first part of Assignment 18 appreared at the beginning of this query. ??i-???????????·?assignment #017
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22:15:11 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 --> First of all this line of questions comes from assignment 18. I don't fully understand, but then again, I don't have my DVD with me. For the first table I have: (0, 0), (1, 2), (2, 8) and (3,18) (0, 0), (1, 1.41) (2, 2.83), (3, 4.24) confidence assessment: 0
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22:15:52 ** 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, so I was right and didn't even know it!!! Cool! self critique assessment: 2
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22:20:14 It the first difference of the `sqrt(y) sequence constant and nonzero?
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RESPONSE --> no confidence assessment: 0
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22:21:32 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 didn't really understand what was being asked. I do however see what I needed to look at to get the answer confidence assessment: 2
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22:28:50 Give your values of m and b for the linear function that models your table.
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RESPONSE --> Use the slope formula (y2-y1)/(x2-x1) (2.8-1.4)/(2-1) 1.4/1 1.4 = m Then sub: 2.8 = 1.4 * 2 + b 2.8 = 2.8 +b 0 = b so the linear funtion is y = 1.4x confidence assessment: 1
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22:29:05 ** 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 --> self critique assessment: 3
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22:30:05 Does the square of this linear functiongive you back the original function?
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RESPONSE --> Yes. confidence assessment: 1
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22:30:33 ** 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 --> Didn't go into the detail I should have self critique assessment: 1
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22:43:38 problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.
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RESPONSE --> The first table gave me: x y 0 7 1 21 2 63 3 189 The second table: x y 0 2.6 1 4.6 2 7.9 3 13.7 Unfortuantely there is not a set pattern making slope difficult to find. I did plug in 7.9-4.6 / 2-1 and got a m=3.3 I substituted that back in and got 21 = 3.3 * 1 + b and got b = 17.7 confidence assessment: 0
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22:46:02 ** 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 --> I did not read where it said to find the log. When I went back and plugged in the numbers for the log I got the same results. self critique assessment: 1
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23:14:30 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 --> They are not equal: First table: x y 0 0 1 5 2 20 3 45 4 80 5 125 Sqrt(y) table: x y 0 0 1 2.2 2 4.5 3 6.7 4 8.9 5 11.2 I used 8.9 - 6.7 / 4 - 3 = 2.2/1 = 2.2 8.9 = 2.2 * 4 + b 8.9 = 8.8 + b .1 = b sqrt(y) = 2.2x + .1 confidence assessment: 1
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23:23:10 ** 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 --> I think I understand what was being asked and got a close answer self critique assessment: 1
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23:55:21 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 will have to reread this section and watch the DVD...something about this is just not registering. confidence assessment: 0
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23:55:41 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 --> self critique assessment: 0
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00:19:29 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 found the log(y): x y .5 -.15 1 -.01 1.5 .02 2 .16 2.5 .19 But once I get here, I am not sure where to go. Again, I will have to look at the DVD. confidence assessment: 0
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00:23:37 ** 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 will have to watch the DVD. Also, I need to reread the log info from Assignment 17, because this skipped it. self critique assessment: 0
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12:11:05 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 --> I got the following tables; x y 0 0 1 1 2 4 Then u w 0 0 1 1 4 2 and x u=f(x)=x^.5 0 0 1 1 2 1.41 confidence assessment: 0
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12:12:05 ** 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 wasn't sure what the ""step of .5"" was. But looking at the explaination shows it. self critique assessment: 2
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12:25:56 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 x^2 function is a smooth parabola that increases, whose vertex is at (0, 0) and would be symmetric with the y axis The sqrt function is a smooth parabola that is increases, whose vertex is at (0,0) but would be symmetric with the x axis. confidence assessment: 1
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12:30:42 ** 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 --> I didn't fully get all of that when I did my graph. I don't understand how the inverse is decreasing when the numbers are .5, 1, 1.5, and 2. I understand where to two curves meet. I didn't plot this on the graph provided which is why I didn't know much about the y=x line. self critique assessment: 1
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12:45:05 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: 1
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12:46:14 ** 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 --> I was expecting numbers here. This doesn't give detatils, only what we have already talked about by the square root being the inverse. self critique assessment: 1
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12:49:08 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 at that rate, every number would have to be squared and if any number was missing from the second column, it would mean that its square root was missing from the first. This can't be possible because the first column is supposed to contain all numbers from 0 to infinity, and the square root of every number from 0 to infinity would be present. confidence assessment: 3
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12:49:20 ** 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 --> self critique assessment: 3
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12:49:57 What number would appear in the second column next to the number 4.31 in the first column?
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RESPONSE --> the square root of 4.31 is 2.08 confidence assessment: 2
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12:50:43 ** 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 did square root insead of squaring the number, but understand where the number came from. self critique assessment: 1
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12:51:17 What number would appear in the second column next to the number `sqrt(18) in the first column?
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RESPONSE --> THIS is where I can use the sqrt... which is 4.24 confidence assessment: 2
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12:53:14 ** The square of sqrt(18) is 18, so 18 would appear in the second column. **
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RESPONSE --> ???? How could the sqare root of 18 be 18? I am slightly confused. self critique assessment: 1
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12:53:36 What number would appear in the second column next to the number `pi in the first column?
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RESPONSE --> pi^2 confidence assessment: 1
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12:53:48 ** pi^2 would appear in the second column. **
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RESPONSE --> FINALLY I got one! self critique assessment: 3
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12:55:26 What would we obtain if we reversed the columns of this table?
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RESPONSE --> the sqrt function where the answer to the previous function is x and the original number is y. confidence assessment: 0
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12:55:56 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 --> I didn't say it quite as well, but I was trying to say the same thing. self critique assessment: 3
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12:56:28 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 --> Now the answer I gave earlier comes into play: 2.08 confidence assessment: 3
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12:56:33 ** you would have sqrt(4.31) = 2.076 **
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RESPONSE --> self critique assessment: 3
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12:56:47 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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12:56:53 ** 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 --> self critique assessment: 3
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12:57:52 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 --> There would not be a number....-3 is not iin the 0-infinity range! Plus you can't take the sqrt of a negative number confidence assessment: 1
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12:57:58 ** -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 --> self critique assessment: 3
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13:03:26 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18
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RESPONSE --> x = log(base 2)(18) x = log(18)/log(2) x = 1.255/.301 x = 4.17 confidence assessment: 2
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13:03:49 ** 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 --> It said solve if possible! self critique assessment: 3
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13:05:07 2 ^ (4x) = 12
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RESPONSE --> 4x = log(base 2)(12) 4x = log (12)/log(2) confidence assessment: 2
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13:05:13 ** 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: 3
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13:06:17 5 * 2^x = 52
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RESPONSE --> First divide by 5 2^x = 10.4 x = log (base 2)(10.4) x = log(10.4) / log(2) confidence assessment: 1
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13:06:26 ** 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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13:07:03 2^(3x - 4) = 9.
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RESPONSE --> (3x-4) = log(9)/(log2) confidence assessment: 1
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13:07:40 ** 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 --> The others didn't go this far so that is why I didn't either, but I do understand. self critique assessment: 2
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13:12:22 14. Solve each of the following equations: 2^(3x-5) + 4 = 0
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RESPONSE --> 2^(3x-5) + 4 = 0 Subtract 4 from both sides: 2^(3x-5) = -4 Then: 3x - 5 = log (-4) / log(2) 3x = log (-4) / log (2) + 5 x = (log -4/ log 2 +5) / 3 But this can't be done! You can't take a log of a negative # confidence assessment: 1
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13:12:29 ** 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 --> self critique assessment: 3
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13:13:56 2^(1/x) - 3 = 0
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RESPONSE --> 2^(1/x) - 3 = 0 2^(1/x) = 3 1/x = log 3 / log 2 1/x = (log 3 / log 2) confidence assessment: 1
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13:14:33 ** 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 --> I just wasn't sure where to go once I got to 1/x. I never thought to inverse the others. self critique assessment: 1
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13:16:26 2^x * 2^(1/x) = 15
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RESPONSE --> 2^x * 2^(1/x) = 15 2 ^ (1/x +x) = 15 1/x +x = log 15/ log 2 confidence assessment: 1
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13:17:20 ** 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 --> Ok, understand how to do this one. Again, I wasn't sure exactly where to go once I got to x + 1/x... self critique assessment: 1
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13:19:53 (2^x)^4 = 5
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RESPONSE --> (2^x)^4 = 5 This would be the same as 2^4x = 5 SO: 4x = log 5 / log 2 x = (log 5 / log 2) /4 x = (.699 / .301) /4 x = 2.322/4 x = .581 confidence assessment: 1
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13:21:27 ** 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 --> It ended up being the same answer! self critique assessment: 3
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