course Mth163

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

017. `query 17

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Question: `qLinearizing 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:

for t =0, 1, 2, 3 y = 0, 2, 8, 18

for t = 0, 1, 2, 3 `sqrt y = 0, 1.4, 2.8, 4.2

Confidence rating: positive

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

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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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Self-critique (if necessary):

ok

Self-critique rating:

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

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Self-critique (if necessary):

The differences of the `sqrt y is constant an nonzero because the first difference is 1.4 and between the next two, 4.2 – 2.8 = 1.4

Self-critique rating: positive.

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Question: `qThe 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

Self-critique rating:

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

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Your solution:

The slope is 1.4 because it is linear and constant, I will plug in one set of coordinates, (2, 2.8) to find b:

2.8 = 1.4 * 2 + b….2.8 = 2.8 + b ….. b = 0

You linear function for this model is: y = 1.4 x + 0

Confidence rating: positive

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

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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):

model is better represented as: `sqrt(y) = 1.4 t

Self-critique rating:

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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:

Our function is: `sqrt(y) = 1.4 t to see if it gives back our original function we take the sqrt of each side:

y = (1.4t)^2……y = 1.96 t^2

Or original function was y = 2 t^2 which 1.96 t^2 is not the exact original but will round up to 2.

Confidence rating: positive

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

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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, can say will round to 2 significant figures

Self-critique rating:

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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:

first we have to have a table for the function y = 7(3^t) using values of t = 0, 1, 2, 3

t y

0 7

1 21

2 63

3 189

now we make a table of t vs log(7(3^t))

t log(7(3^t))

o .85

1 1.32

2 1.80

3 2.28

the difference runs .47, .48, .48, giving you an approx slope of .472,

giving you log(y) = .472x + b….solving with x, y coordinates to get b = .85

giving you log(y) = .472t + .85….10^(log y) = 10^(.472t + .85)

y = 10^(.472t) * 10^(.85)….we can do this because of the properties of logs.

y = 2.97^t * 7.08… which is only 2 digits away from our original equation.

Confidence rating: mostly

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

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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

Self-critique rating:

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Question: `qproblem 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:

t = 0, 1, 2, 3, 4 using function `sqrt(y) = 2.27x + .05 y = .05, 2.32, 4.59, 6.86,9.13

the square root of these values are: .224, 1.52, 2.14, 2.62,3.02

the graph of these points is nearly a straight line

???? I don’t understand how to do this ????

Confidence rating: unsure

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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.**

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

INSTRUCTOR RESPONSE: If your points lie along or near a straight line then the straight line is of the form y = m x + b.

b is the y intercept of the straight line, m is the slope. For the given data, after linearizing we found that m was about 1.23.

If the line happens to go through the origin then the y intercept is 0, so b = 0.

Otherwise b is not zero.

You simply accept whatever the 'best' straight line tells you. Sometimes b will be zero, but usually it won't.

In the current example, it happens that the 'best' straight line goes through the vertical axis at b = .05.

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Self-critique (if necessary):

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

Self-critique rating:

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Question: `qproblem 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:

using y = 2^x

x y

4 16

5 25

6 36

7 49

8 64

x log(y)

4 1.20

5 1.40

6 1.56

7 1.69

8 1.81

giving you the best fit to this data as log(y) = .2124x + 3.79

solving this equation we get:

10^log(y) = 10^(.2124x + 3.79)

y = (10^.2124)^x * (10^3.79)

y = 1.63^x * 6166

Confidence rating: unsure

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

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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, if I did my calculations right.

Remember to check with original data and show that the model is good by its slim deviations.

Self-critique rating:

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Question: `qproblem 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:

Confidence rating:

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

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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. **

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Self-critique (if necessary):

&#This also requires a self-critique.

&#

Self-critique rating:

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Question: `qInverse 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:

x f(x)

0 0

.5 .25

1 1

1.5 2.25

2 4

for the inverse:

x f^-1(x)

0 0

.25 .5

1 1

2.25 1.5

4 2

Confidence rating: positive

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

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Given Solution:

`a** The table is

x f(x)

0 0

.5 .25

1 1

1.5 2.25

2 4.

Reversing columns we get the following partial table for the inverse function:

x f^-1(x)

0 0

.25 .5

1 1

2.25 1.5

4 2 **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qDescribe 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 y = x^2 curves upward increasing at an increasing rate and the inverse to this line is increasing at a decreasing rate.

They share two pair of coordinates, (0,0) and (1,1) and is a reflection of each other in respective of the y = x line.

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `q8. 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:

you would have a table with y = x^2 and y = square root(x)

Confidence rating: positive

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

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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):

Remember functions are for x > = 0

Self-critique rating:

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Question: `q9. 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:

if the first column consist of all #’s possible and the second column is of all numbers squared then it is s=certain that the numbers from the first column will appear once in the second column.

Confidence rating: positive

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

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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):

Note: the reason a number cannot appear twice in the second column is because its square root would have to appear twice in the first column.

Self-critique rating:

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Question: `qWhat number would appear in the second column next to the number 4.31 in the first column?

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Your solution:

18.58 approx.

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `qWhat number would appear in the second column next to the number `sqrt(18) in the first column?

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Your solution:

18

Confidence rating: positive

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

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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

Self-critique rating:

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

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Your solution:

pi^2

Confidence rating: mostly

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

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Given Solution:

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

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Self-critique (if necessary):

ok

Self-critique rating:

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

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Your solution:

our tables would be the opposite of the first set:

the first column would consist of the #’s squared and the second column would consist of the square roots of the first column.

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `qWhat 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:

approx. 2.08

Confidence rating: positive

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

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Given Solution:

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

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat 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

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `qWhat 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:

nothing because -3 isn’t a real number and you can’t square a negative so it wouldn’t be in the second column.

-3 is a real number but its square root isn't.

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `q13. Translate each of the following exponential equations into equations involving logarithms, and

solve where possible: 2 ^ x = 18

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Your solution:

2^x = 18

log{base 2}(2^x) = log{base2}(18)

x = log{base2}(18)

x = log18/ log 2

x = 4.17 approx.

Confidence rating: positive

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

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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

Self-critique rating:

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Question: `q2 ^ (4x) = 12

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Your solution:

2^(4x) = 12

I would simplify first (2^4)^x = 12……16^x = 12

log{base16}(16^x) = log{base 16}(12)

x = log12 / log 16

Confidence rating: positive

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

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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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Self-critique (if necessary):

???? 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 ????

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

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

Self-critique rating:

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Question: `q5 * 2^x = 52

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Your solution:

5 * 2^x = 52 first you need to simplify it some:

2^x = 52/5

log{base2}(2^x) = log{base2}(52/5)

x = log (52/5) / log 2

Confidence rating: mostly

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

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

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

Self-critique rating:

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Question: `q2^(3x - 4) = 9.

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Your solution:

2^(3x – 4) = 9

(3x – 4) log(2) = log (9)

3x – 4 = log (9) / log(2)

3x = log9 / log 2 + 4

x =[ log 9 / log 2 + 4] / 3

Confidence rating: unsure

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

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

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

Self-critique rating:

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Question: `q14. Solve each of the following equations:

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

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Your solution:

2^(3x – 5) = -4

(3x – 5) log (2) = log (-4)

3x – 5 = log -4 / log 2

3x = log -4 / log 2 + 5

x = (log -4 / log 2 + 5) / 3

Confidence rating: positive

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

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

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 there is no solution because there is no log of a negative number.***

Self-critique rating:

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Question: `q2^(1/x) - 3 = 0

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Your solution:

2^(1/x) – 3 = 0

2^(1/x) = 3

(1/x) log(2) = log (3)

1 / x = log 3 / log 2

x = log 3 / log 2

Confidence rating: positive

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

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

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

Self-critique rating:

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Question: `q2^x * 2^(1/x) = 15

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Your solution:

2^x * 2 ^(1/x) = 15

rules of logarithms allows you to rewrite this as:

2^( x + 1/x) = 15

x + 1/x = log 15 / log 2

x^2 + 1 = log 15 / log 2

To get that last step you would have multiplied the preceding step by x. This would have given you

x^2 + 1 = log 15 / log 2 * x.

So you're stuck with the quadratic formula.

x^2 = log 15 / log2 - 1

x = (log 15 / log 2 – 1)^1/2

Confidence rating: unsure

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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. **

Self-critique (if necessary)

:

ok, I understand were you have the quadratic formula but I don’t understand where you have gotten c as 4.????

???? Is the way I solved my problem wrong or is that a way that it can be done ????

I did have c = 4. Unlikely that was a simple typo; more likely I was thinking ahead to the discriminant term -4 a c.

I've changed the value of c in the above.

The solutions are still correct.

Self-critique rating:

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

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Your solution:

(2^x)^4 = 5…first get rid of the ^4 and move it to the other side.

(2^x) = 5 ^(1/4)

log{base2}(2x) = ¼ log{base2}(5)

x = ¼ log 5 / log 2

Confidence rating: positive

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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):

ok

Self-critique rating:

"

&#Good responses. See my notes and let me know if you have questions. &#