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course Mth 163

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Question: `q 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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Your solution:

Y vs t is 0,0;1,2;2,8;3,18 sqrtIy) vs t, with sqrt(y) is 0,0;1,1.4;2,2.8;3,4.3

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Given Solution:

`a** 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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Question: `q It the first difference of the `sqrt(y) sequence constant and nonzero?

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Your solution: Non zero

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Question: `q 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.

You solution: constant

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Question: `q Give your values of m and b for the linear function that models your table.

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Your solution: 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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Given Solution:

`a** 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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Question: `q Does the square of this linear function give you back the original function?

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Your solution: Yes

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3

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Given Solution:

`a** 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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Question: `q problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The answer is identical to the figure given.

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3

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Given Solution:

`a** 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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Question: `q problem 7. Hypothesized fit is `sqrt(y) = 2.27 x + .05.

Compare your result to the 'ideal' y = 5 t^2 function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 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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3

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Given Solution:

`a** 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付 + 0.27. Your function should be reasonably close to this but will probably not be identical.

Squaring both sides we get y = 5.1529付^2 + 1.2258付 + 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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Question: `q 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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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Question: `q 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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Your solution: 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.

he 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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Question: `q 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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Your solution:

The table is

x y = f(x)

0 0

0.5 0.25

1 1

1.5 2.25

2 4

table for the inverse function:

x f ^ -1 (x)

0 0

0.25 0.5

1 1

2.25 1.5

4 2

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Question: `q 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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Your solution:

The curve of the original function is growing at an increasing rate, the curve for the inverse function is accumulative at a reducing rate. The curves meet at (0, 0) and at (1, 1).The line joining the pairs of points passes through the y = x line at a right angle, and the y = x line intersects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The inverted table would give us the table for the square root function y = sqrt(x) The y = x^2 and y = sqrt(x)functions are opposite functions for x >= 0

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The second column involves of all the squares. For a number to appear in the second column the square root of that number would have to be in the first. Meanwhile every possible number appears in the first column, then no matter what number we pick it will appear in the second column. So each possible positive number appears in the second column. If a number appears double in the second column then its square root would appear twice in the first column

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number 4.31 in the first column?

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Your solution:

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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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number `sqrt(18) in the first column?

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Your solution:

18

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Given Solution:

`a** The square of sqrt(18) is 18, so 18 would appear in the second column. **

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Question: `q What number would appear in the second column next to the number `pi in the first column?

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Your solution:

Pi^2

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3

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Given Solution:

`a** pi^2 would appear in the second column. **

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Question: `q What would we obtain if we reversed the columns of this table?

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Your solution:

The table would have the square of the 2nd column in the first, so the 2nd column would be the square of the first.

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Given Solution:

`aOur 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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Question: `q 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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Your solution:

2.076

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3

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Given Solution:

a** you would have sqrt(4.31) = 2.076 **

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Question: `q 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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Your solution:

pi

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number -3 in the first column of this table?

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Your solution:

-3

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Given Solution:

`a** -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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Question: `q 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18

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Your solution:

b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2)

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Given Solution:

`a** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2). **

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Question: `q 2 ^ (4x) = 12

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Your solution:

4 x =log{base 2}(12) = log(12) / log(2)

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2

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Given Solution:

`a** 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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Question: `q 5 * 2^x = 52

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Your solution:

2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2)

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Given Solution:

`a** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **

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Question: `q 2^(3x - 4) = 9.

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Your solution:

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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Given Solution:

`a** 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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Question: `q 14. Solve each of the following equations:

2^(3x-5) + 4 = 0

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Your solution:

No solution

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3

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Given Solution:

`a** 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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Question: `q 2^(1/x) - 3 = 0

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Your solution:

.63

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Given Solution:

`a** 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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Question: `q 2^x * 2^(1/x) = 15

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Your solution:

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Given Solution:

a** 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 = 0then use quadratic formula with a=1, b=-log{base 2}(15) and c=1. Our solutions are x = 0.2753664762 OR x = 3.631524119. **

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Question: `q (2^x)^4 = 5

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Your solution:

2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2)

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Given Solution:

`a** 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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Self-critique (if necessary):

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Self-critique rating:

#$&*

course Mth 163

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Question: `q 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Y vs t is 0,0;1,2;2,8;3,18 sqrtIy) vs t, with sqrt(y) is 0,0;1,1.4;2,2.8;3,4.3

confidence rating #$&*:

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3

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Given Solution:

`a** 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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Question: `q It the first difference of the `sqrt(y) sequence constant and nonzero?

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Your solution: Non zero

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Question: `q 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.

You solution: constant

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Question: `q Give your values of m and b for the linear function that models your table.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 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.

confidence rating #$&*:

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2

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Given Solution:

`a** 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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Question: `q Does the square of this linear function give you back the original function?

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Your solution: Yes

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3

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Given Solution:

`a** 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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Question: `q problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: The answer is identical to the figure given.

confidence rating #$&*:

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3

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Given Solution:

`a** 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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Question: `q problem 7. Hypothesized fit is `sqrt(y) = 2.27 x + .05.

Compare your result to the 'ideal' y = 5 t^2 function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 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

confidence rating #$&*:

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3

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Given Solution:

`a** 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付 + 0.27. Your function should be reasonably close to this but will probably not be identical.

Squaring both sides we get y = 5.1529付^2 + 1.2258付 + 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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Question: `q 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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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Question: `q 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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Your solution: 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.

he 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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Question: `q 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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Your solution:

The table is

x y = f(x)

0 0

0.5 0.25

1 1

1.5 2.25

2 4

table for the inverse function:

x f ^ -1 (x)

0 0

0.25 0.5

1 1

2.25 1.5

4 2

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Question: `q 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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Your solution:

The curve of the original function is growing at an increasing rate, the curve for the inverse function is accumulative at a reducing rate. The curves meet at (0, 0) and at (1, 1).The line joining the pairs of points passes through the y = x line at a right angle, and the y = x line intersects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The inverted table would give us the table for the square root function y = sqrt(x) The y = x^2 and y = sqrt(x)functions are opposite functions for x >= 0

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The second column involves of all the squares. For a number to appear in the second column the square root of that number would have to be in the first. Meanwhile every possible number appears in the first column, then no matter what number we pick it will appear in the second column. So each possible positive number appears in the second column. If a number appears double in the second column then its square root would appear twice in the first column

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number 4.31 in the first column?

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Your solution:

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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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number `sqrt(18) in the first column?

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Your solution:

18

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Given Solution:

`a** The square of sqrt(18) is 18, so 18 would appear in the second column. **

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Question: `q What number would appear in the second column next to the number `pi in the first column?

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Your solution:

Pi^2

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Given Solution:

`a** pi^2 would appear in the second column. **

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Question: `q What would we obtain if we reversed the columns of this table?

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Your solution:

The table would have the square of the 2nd column in the first, so the 2nd column would be the square of the first.

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Given Solution:

`aOur 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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Question: `q 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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Your solution:

2.076

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Given Solution:

a** you would have sqrt(4.31) = 2.076 **

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Question: `q 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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Your solution:

pi

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number -3 in the first column of this table?

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Your solution:

-3

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Given Solution:

`a** -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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Question: `q 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18

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Your solution:

b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2)

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Given Solution:

`a** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2). **

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Question: `q 2 ^ (4x) = 12

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Your solution:

4 x =log{base 2}(12) = log(12) / log(2)

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Given Solution:

`a** 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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Question: `q 5 * 2^x = 52

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Your solution:

2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2)

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Given Solution:

`a** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **

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Question: `q 2^(3x - 4) = 9.

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Your solution:

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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Given Solution:

`a** 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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Question: `q 14. Solve each of the following equations:

2^(3x-5) + 4 = 0

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Your solution:

No solution

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Given Solution:

`a** 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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Question: `q 2^(1/x) - 3 = 0

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Your solution:

.63

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Given Solution:

`a** 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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Question: `q 2^x * 2^(1/x) = 15

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Your solution:

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Given Solution:

a** 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 = 0then use quadratic formula with a=1, b=-log{base 2}(15) and c=1. Our solutions are x = 0.2753664762 OR x = 3.631524119. **

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Question: `q (2^x)^4 = 5

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Your solution:

2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2)

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Given Solution:

`a** 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). **

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

#$&*

course Mth 163

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Question: `q 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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Your solution:

Y vs t is 0,0;1,2;2,8;3,18 sqrtIy) vs t, with sqrt(y) is 0,0;1,1.4;2,2.8;3,4.3

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Given Solution:

`a** 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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Question: `q It the first difference of the `sqrt(y) sequence constant and nonzero?

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Your solution: Non zero

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Question: `q 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.

You solution: constant

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Question: `q Give your values of m and b for the linear function that models your table.

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Your solution: 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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Given Solution:

`a** 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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Question: `q Does the square of this linear function give you back the original function?

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Your solution: Yes

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Given Solution:

`a** 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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Question: `q problem 2. Linearize the exponential function y = 7 (3 ^ t). Give your solution to the problem.

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Your solution: The answer is identical to the figure given.

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Given Solution:

`a** 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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Question: `q 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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Your solution: 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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Given Solution:

`a** 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付 + 0.27. Your function should be reasonably close to this but will probably not be identical.

Squaring both sides we get y = 5.1529付^2 + 1.2258付 + 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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Question: `q 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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Your solution:

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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Question: `q 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution: 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.

he 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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Question: `q 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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Your solution:

The table is

x y = f(x)

0 0

0.5 0.25

1 1

1.5 2.25

2 4

table for the inverse function:

x f ^ -1 (x)

0 0

0.25 0.5

1 1

2.25 1.5

4 2

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Question: `q 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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Your solution:

The curve of the original function is growing at an increasing rate, the curve for the inverse function is accumulative at a reducing rate. The curves meet at (0, 0) and at (1, 1).The line joining the pairs of points passes through the y = x line at a right angle, and the y = x line intersects each connecting line. So the two graphs are mirror images of one another with respect to the line y = x

confidence rating #$&*:

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The inverted table would give us the table for the square root function y = sqrt(x) The y = x^2 and y = sqrt(x)functions are opposite functions for x >= 0

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Given Solution:

`a** 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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Question: `q 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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Your solution:

The second column involves of all the squares. For a number to appear in the second column the square root of that number would have to be in the first. Meanwhile every possible number appears in the first column, then no matter what number we pick it will appear in the second column. So each possible positive number appears in the second column. If a number appears double in the second column then its square root would appear twice in the first column

confidence rating #$&*:

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number 4.31 in the first column?

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Your solution:

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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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number `sqrt(18) in the first column?

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Your solution:

18

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Given Solution:

`a** The square of sqrt(18) is 18, so 18 would appear in the second column. **

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Question: `q What number would appear in the second column next to the number `pi in the first column?

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Your solution:

Pi^2

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Given Solution:

`a** pi^2 would appear in the second column. **

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Question: `q What would we obtain if we reversed the columns of this table?

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Your solution:

The table would have the square of the 2nd column in the first, so the 2nd column would be the square of the first.

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Given Solution:

`aOur 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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Question: `q 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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Your solution:

2.076

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Given Solution:

a** you would have sqrt(4.31) = 2.076 **

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Question: `q 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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Your solution:

pi

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3

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Given Solution:

`a** 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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Question: `q What number would appear in the second column next to the number -3 in the first column of this table?

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Your solution:

-3

confidence rating #$&*:

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3

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Given Solution:

`a** -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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Question: `q 13. Translate each of the following exponential equations into equations involving logarithms, and solve where possible: 2 ^ x = 18

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Your solution:

b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2)

confidence rating #$&*:

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2

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Given Solution:

`a** b^x = a is expressed in logarithmic form as x = log{base b}(a) so this equation would be translated as x = log{base2}(18) = log(18) / log(2). **

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Question: `q 2 ^ (4x) = 12

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Your solution:

4 x =log{base 2}(12) = log(12) / log(2)

confidence rating #$&*:

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2

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Given Solution:

`a** 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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Question: `q 5 * 2^x = 52

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Your solution:

2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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Given Solution:

`a** You get 2^x = 52/5 so that x = log{base 2}(52/5) = log(52/5) / log(2). **

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Question: `q 2^(3x - 4) = 9.

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Your solution:

3x - 4 = log 9 / log 2 so that 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3

confidence rating #$&*:

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3

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Given Solution:

`a** 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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Question: `q 14. Solve each of the following equations:

2^(3x-5) + 4 = 0

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Your solution:

No solution

confidence rating #$&*:

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3

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Given Solution:

`a** 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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Question: `q 2^(1/x) - 3 = 0

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Your solution:

.63

confidence rating #$&*:

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3

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Given Solution:

`a** 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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Question: `q 2^x * 2^(1/x) = 15

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Your solution:

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

a** 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 = 0then use quadratic formula with a=1, b=-log{base 2}(15) and c=1. Our solutions are x = 0.2753664762 OR x = 3.631524119. **

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Question: `q (2^x)^4 = 5

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Your solution:

2^x = 5^(1/4) so that x = log(5^(1/4) ) / log(2) = 1/4 log(5) / log(2)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

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Given Solution:

`a** 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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