Assignment 18

#$&*

course Mth 163

3/30/13 around 11 p.m.

018. `query 18

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Question: `q Explain the general strategy used in this assignment for attempting to linearize a set of y vs. x data.

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Your solution:

If the y vs. x table shows a linear graph you just have to find the slope and y-intercept. The equation will be y = m x + b.

If your table isn’t a linear graph then you have to check the graphs of log(y) vs. x, y vs. log(x) and log(y) vs. log(x) to see if any of these tables result in a linear graph.

confidence rating #$&*: 3

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Given Solution:

`a** If the y vs. x table yields a decent linear graph, you'll just find the slope and vertical-axis intercept and you're pretty much done. Your equation in this case will be

y = m x + b.

If your table doesn't yield a linear graph, then you'll check the graphs of log(y) vs. x, y vs. log(x) and log(y) vs. log(x) to see if any of these tables yields a linear graph.

You won't have set the system up the same way it's presented here, but your setup should be consistent with the following:

Start with your table of y vs. x values. Your table might have headings

x y

with values listed in the columns below.

Add columns for log(x) and log(y), so your headings now read

x y log(x) log(y)

You can easily fill in the values of log(x) and log(y). Use can use the base-10 log, the natural log, or the log to any other base. Usually a special base won't be helpful to you so you will keep it simple by using base-10 or natural log.

Now you can just copy the columns to create the necessary tables. The first table might be log(y) vs. log(x), in which case it would have headings

log(x) log(y).

The second might be log(y) vs. x, with headings

x log(y).

The third might be y vs. log(x), with headings

log(x) y.

Having filled in the columns to create each table, you will want to see which table, if any, gives you a linear graph. Hopefully you are able to do some estimates using mental arithmetic to quickly eliminate one or more possibilities.

You then graph any of the possibilities you haven't eliminated. (If your mental arithmetic isn't reliable you'll probably end up having to graph all possibilities).

Any graph which can be reasonably well fit with a straight line will give you a linearization. Finding the slope m and the vertical-axis intercept b you will obtain an equation for the line.

If the log(y) vs. x graph is linear, then the corresponding equation is

log(y) = m x + b.

To express y as a function of x, this equation would then be solved for y.

If the log(y) vs. log(x) graph is linear, then the corresponding equation is

log(y) = m log(x) + b.

To express y as a function of x, this equation would then be solved for y.

If the y vs. log(x) graph is linear, then the corresponding equation is

y = m log(x) + b.

If none of these graphs is linear, then you would have to get creative and consider other possibilities. However, in this course it will be sufficient if you simply apply this method correctly.

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Self-critique (if necessary): I did not write all of this out, but I understand all of the steps and would be able to do so if needed

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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:

The table for y vs. t is

t y

0 0

1 2

2 8

3 18

The table for sqrt(y) vs t, with sqrt(y) is

t sqrt(y)

0 0

1 1.4

2 2.8

3 4.2

confidence rating #$&*: 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 sqrt(y) vs t, with sqrt(y) given to 2 significant figures, is

t sqrt(y)

0 0

1 1.4

2 2.8

3 4.2 **

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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

My values were (t, sqrt(y) ) are (0,0), (1, 1.4), (2, 2.8), (3, 4.2). These points fit a straight line through the origin with slope 1.4, so the equation of the line is sqrt(y) = 1.4 t + 0, or sqrt(y) = 1.4 t

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 , the original function was y = 2 t^2. Squaring both sides of sqrt(y) = 1.4 t we get y = 1.96 t^2.

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

My 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 about .473.

log(y) is a linear function of t with slope .473 and y- intercept .85.

Therefore we have log(y) = .473 t + .85. Thus

10^(log y) = 10^(.473 t + .85) so

y = 10^(.473 t) * 10^(.85) or

y = (10^.473)^t * (10^.85

y = 2.97^t * 7.08. or y = 3 * 7^t.

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

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 get a straight-line

The best-fit linear function to sqrt(y) vs. x gives us sqrt(y) = 2.32•t + 0.24.

Squaring both sides we get y = 5.3824•t^2 + 1.2258•t + 0.0576.

If 0.0576 is neglected we get y = 5.38 t^2 + 1.23 t.

Because of the 1.23 t term this isn't a good approximation we get y = 5 t^2.**

confidence rating #$&*: 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•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.**

BRIEF SUMMARY:

If `sqrt(y) = 2.27 x + .05, then squaring both sides gives us

y = (2.27 x + .05)^2,

which when expanded gives you something fairly close to y = 5 t^2, but not all that close.

STUDENT COMMENT ok i dont really understand where the 1.23 came from in the first place

INSTRUCTOR RESPONSE:

The 1.23 arises when you square the binomial.

(a + b)^2 = a^2 + 2 a b + b^2. So

(2.27 x + .05)^2 = 5.15 t^2 + 1.23 t + 0.073

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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) 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 =-.00078.

confidence rating #$&*: 3

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Given Solution:

`aFor (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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 we get

10^log(y) =10^( 0.5074 log(x) - 0.0056)

This simplifies to: 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, and when we solve for y we get:

10^log(y) = 10^(0.1735x + 0.0122) or

y = 10^.0122 * 10^(0.1735•x) = 1.0285 * 1.491^x.

confidence rating #$&*: 3

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

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Given Solution:

`a** 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. **

STUDENT QUESTION:

Im not sure how to come up with this

10^log(y) =10^( 0.5074 log(x) - 0.0056)

which we simplify to obtain

y = 0.987•x^0.507.

INSTRUCTOR RESPONSE:

I assume you understand why log(y) vs. x is linear, while log(y) vs. log(x) is not, and why log(y) = 0.1735x + 0.0122.

So solve log(y) = q we would use the fact that log(y) and 10^y are inverse functions, so that 10^(log(y)) = y. The equation

log(y) = q implies the equation 10^(log(y)) = 10^q, which becomes y = 10^q.

We apply the same strategy to the present equation.

log(y) = 0.1735x + 0.0122 becomes

10^(log(y)) = 10^(0.1735 x + 0.0122), which becomes

y = 10^(0.1735 x + 0.0122). Applying the laws of exponents this becomes

y = 10^(0.1735 x) * 10^0.0122. Since 10^(0.1735 x) = (10^0.1735) ^x = 1.491^x, and 10^0.0122 = 1.0285, we have

y = 1.0285 * 1.491^x .

10^log(y) =10^( 0.5074 log(x) - 0.0056) becomes y = 0.987•x^0.507 by a very similar series of steps:

10^log(y) = y

10^(0.5074 log(x)) = (10^(log(x)) )^ 0.5074 = x ^ 0.5074

10^-0.0122 = 0.987

so the equation

10^log(y) =10^( 0.5074 log(x) - 0.0056) becomes

10^(log y) = 10^(.5074 log(x)) * 10^-0.0056, which in turn becomes

y = 0.987•x^0.507.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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

Reversing columns we get:

x f ^ -1 (x)

0 0

0.25 0.5

1 1

2.25 1.5

4 2

confidence rating #$&*: 3

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

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Given Solution:

`a** The table is

x y = f(x)

0 0

0.5 0.25

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

0.25 0.5

1 1

2.25 1.5

4 2

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 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.

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok 3

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Self-critique rating: 3

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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 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.

confidence rating #$&*: 3

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

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 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.

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

4.31^2 = 18.5761 would appear in the second column

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

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

confidence rating #$&*: 3

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

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 would appear in the second column

confidence rating #$&*: 3

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

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Given Solution:

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

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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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Your 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

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

sqrt(4.31) = 2.076

confidence rating #$&*: 3

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

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Given Solution:

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

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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:

Since pi^2 is in the first column u would take the squr root which would be pi

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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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 couldn’t appear in the first column so doesn’t have an answer

confidence rating #$&*: 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. **

STUDENT COMMENT: (student gave the answer 1.73 i)

oh wow that was really tricky

INSTRUCTOR RESPONSE: sqrt(-3) = 1.73 i is a very good answer; if the domain and range of the function include the complex numbers, this would in fact be a 3-significant-figure approximation of the number corresponding to -3.

Since we're dealing here with the real numbers, though, -3 never appears in the second column of the x^2 function, so it won't appear in the first column of the inverse function.

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Self-critique (if necessary): ok

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Self-critique rating: 3

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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{base 2}(18) = log(18) / log(2).

confidence rating #$&*: 3

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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{base

2}(18) = log(18) / log(2). **

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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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

confidence rating #$&*: 3

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

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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). **

STUDENT QUESTION:

Mr.Smith, I don’t understand how you have this laid out. Could give more of a step by step detail so that I might

understand it

INSTRUCTOR RESPONSE:

Sure. Step-by-step:

• b^x = a is expressed in logarithmic form as x = log{base b}(a)

2 ^ (4x) = 12 is of the form b^x = a, but with b = 2, x replace by 4x and a = 12.

Thus the form

x = log{base b}(a)

becomes

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

log{base 2}(12) = log(12) / log(2).

Thus

4x = log(12) / log(2).

You weren't asked to solve this for x, but had you been asked the solution would be found by dividing both sides by 4, which gives us

x = log(12) / (4 log(2) ).

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Self-critique (if necessary): ok

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Self-critique rating: 3

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Question: `q 5 * 2^x = 52

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Your solution:

We would get 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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Your solution:

We would get 3x - 4 = log 9 / log 2 so that 3x = log 9 / log 2 + 4 and x = ( log 9 / log 2 + 4 ) / 3.

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 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:

The solution is impossible because -4 has no solution. We work this out by log(-4)/log(2)=3x - 5.

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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Your solution:

You get 2^(1/x) = 3 so that 1/x = log(3) / log(2) and x = log(2) / log(3) = .63.

confidence rating #$&*: 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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Self-critique (if necessary): ok

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Self-critique rating: 3

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

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Your solution:

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).

confidence rating #$&*: 3

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

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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 = 0

then use quadratic formula with a=1, b=-log{base 2}(15) and c=1.

Our solutions are

x = 0.2753664762 OR x = 3.631524119. **

STUDENT COMMENT

I don't think I would've made the correlation with a quadratic after working with exponential

functions for so long.. I wasn't sure how to combine 1/x + x, either, which I'm guessing should be simple but it's eluding

me at the moment.

INSTRUCTOR RESPONSE

You want to solve the equation, and the most efficient method is to multiply both sides by x.

However to add 1/x + x you put both terms over the common denominator x. You do this by multiplying the second term, which has no denominator, by x / x. We get

(1/x) + x * (x / x) = (1/x) + (x^2 / x) = (1 + x^2) / x.

Again we wouldn't do that here, but that's how it would be done if we wanted to add the two terms.

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Self-critique (if necessary):

I never would have seen the correlation for the quadratic equation either, but I understand based on what the student asked and your answer

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Self-critique rating: 3

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

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Your solution:

We 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).

confidence rating #$&*: 3

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

.............................................

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): ok

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Self-critique rating: 3

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: Show the tables you would get when attempting to linearize the data set

x y

10 3

20 10

30 29

and indicate which is most likely linear. You may use the following approximations: ln(1) = 0, ln(20) = 3.0, ln(10) = 2.3, ln(29) = 3.4 (you haven't been given the value of ln(30) but it should be easy to make a sufficiently accurate commonsense estimate of its value, to within +-0.1 of its actual value, given the information provided).

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Your solution:

The table I get is:

X sqrt

10 1.73

20 3.16

30 5.39

This second graph is mostly likely linear

@&

That table doesn't give you a linear function.

You need to construct the tables for log(y) vs. x, y vs. log(x) and log(y) vs. log(x) and check whether one of them is linear.

*@

confidence rating #$&*: 3

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Question: If the log(y) vs. log(x) table for a function is as follows

log(x) log(y)

1.1 5.0

2.3 9.0

3.9 14.0

what are your estimates of the slope and vertical-axis intercept of the best-fit straight line for the log(y) vs. log(x) graph?

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Your solution:

Y-intercept = + 0.3333

Slope = 4.5

confidence rating #$&*: 3

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Question: If for a certain data set we get ln(y) = 0.31 x + 3, what is resulting y vs. x function, obtained by solving the equation for y?

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Your solution:

Ln (y) = .31x + 3

Ln(y) = ln .31x + 3

@&

The equation you get by taking the natural log of just the right-hand side does not follow from the original equation.

*@

Y = -1.17x + 1.09

@&

ln(a + b) is not the same as ln(a) + ln(b).

*@

Y = 1.09

@&

You did not solve the equation correctly.

*@

confidence rating #$&*: 2

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Question: The table

x y

10 3

20 10

30 29

describes the graph of a certain function y of x. Sketch a quick

What corresponding table describes the y vs. x graph of the inverse function?

Give a brief description of the graph of each function.

Describe in at least two different ways, without reference to the tables, how the graphs of these two functions are related.

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Your solution:

An inverse of the y vs. x graph would be

X Y

3 10

10 20

29 30

The graph of the first function that was given was non-linear with an equation of y = 13x - 12

@&

This equation doesn't fit the data. The slope of your graph will be much less than 13.

You can see this by plotting your equation on the same graph as your data.

*@

@&

The data are not in any case linear. A linear function would not be a good fit.

*@

The graph of the inverse function was the opposite of the first graph that was also non-linear with an equation of y = 0.7182x + 9.9448

@&

The graphs of inverse functions as symmetric about the line y = x. Is that the case for your graphs? Why must it be so for inverse functions?

*@

confidence rating #$&*: 2

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Question: For the function of the previous problem, according to your graph, what would be the approximate y value corresponding to x = 15?

For the inverse function, what approximate y value would correspond to x = 15?

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Your solution:

For the first function we would use the equation obtained y = 13x - 12 and substitute 15 in for x giving us

Y = 13(15) -12

Y = 193

So we have (15, 193)

@&

For the given function x = 15 is between x = 10 and x = 20. For x = 10 and x = 20 the y values are respectively 3 and 10. So your answer 193 doesn't fit the given information.

Had your function model been valid for the data, however, the method you used in this question would have worked.

*@

For the inverse graph we have the equation y = 0.7182x + 9.9448. Substituting 15 in for x in this equation gives us:

Y = 0.7182(15) + 9.9448

Y = 20.7178

So we have (15, 20.7178)

@&

As x changes from 10 to 29 the value of the inverse function changes from 20 to 30. While the result 20.7 isn't terribly wrong, it really doesn't fit the trend of the data.

*@

@&

You should base your estimates for this question and the preceding on the graphs you sketched.

*@

confidence rating #$&*: 3

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Question: What is the exact solution to the equation 3^(5 x + 1) = 20? (For reference on the possible form of an exact solution, the exact solution to the equation 2 * 5^x = 9 is x = (log(9) - log(2)) / log(5) ).

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Your solution:

We start with 3^(5 x + 1) = 20

We take ln of both sides giving us

Ln 3^(5 x + 1) = ln(20) then

@&

ln(3^(5x+1)) = 3 ln(5x+1), not 3 ( 5x + 1 ).

*@

3 (5x + 1) = ln 20

15x + 3 = ln20

15x + 3 = 2.99573

@&

2.99573 is not exact. The question asked for the exact solution, not an approximation.

*@

15x = -.0043

X = -2.85x10^-4

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Self-critique rating:2

@&

You should revise the problems at the end, nearly all of which contained important errors.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&#

*@