qa 17

course MTH 163

11/29/09 3:30 p.m.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

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. `q uery 17

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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:

t y

3 18

2 8

1 2

0 0

sqrty vs t

t sqrty

3 4.24264

2 2.82843

1 1.41421

0 0

This data is linear because the diffrence in the y value is now equal to 1.41421 for every movement of time across the x coordinate

confidence rating: 4

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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?

yes that is the change in y which is 1.41421 which is constant

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

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Your solution:

the 1.41421 would be the slope or m value

b would be 0 the y intercept

confidence rating: 4

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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 functiongive you back the original function?

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Your solution: yes it will

sqrt(y)= 1.41421t

y = 2t^2

confidence rating: 4

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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:

y= 7*3^t

t y

0 7

1 21

2 63

3 189

log(y) vs t

t log(y)

0 .85

1 1.32

2 1.80

3 2.28

change in log(y)~.47

this data is now linear

log(y)=.47t * .85

y= 10^.47(t) + 10^.85

y = 2.95t +7.08

confidence rating: 2

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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:

x y

0 .14

1 5.66

2 23.2

3 52

4 82.2

5 135.5

x sqrt(y)

0 .3742

1 2.3791

2 4.8166

3 7.2111

4 9.0664

5 11.6404

Line of best fit

sqrt(y)= 2.25x +.287

taking the square of both sides

y =5.0625x^2 + .0824

This is relatively close to the ideal function.

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.**

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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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:

t y

0 .42

1 .29

2 .21

3 .15

4 .10

5 .07

log(y) vs t

t log (y)

0 -.377

1 -.538

2 -.678

3 -.824

4 - 1

5 -1.155

line of best fit

log(y)=-.155t + -.375

y= 10^-.155t + 10^-.375

y= .6998t + .4217

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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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:

x y

.5 .7

1 .97

1.5 1.21

2 1.43

2.5 1.56

log(x) Log(y)

-.30103 -.1549

0 -.013228

.176091 .0828

.30103 .1553

.39794 .193125

confidence rating: 1

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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):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. ""

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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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:

f(x)=.5x^2

x f(x)

0 0

1 .5

2 2

x f^-1(x)

0 0

.5 1

2 2

confidence rating: 1

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Given Solution:

`a*

`a** The table is

 

x

y = f(x)

0

0

0.5

0.5

1

2

1.5

4.5

2

8

 

Reversing columns we get the following partial table for the

inverse function:

 

x

f ^ -1 (x)

0

0

0.5

0.5

2

1

4.5

1.5

8

2

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Self-critique (if necessary):

I'm not sure what I did wrong on this one

The table in the given solution was incorrect. You got it right.

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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:

They mirror the y=x and are perpendicular forming right angles

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:

confidence rating:

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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:

All possible numbers appear in the second column

confidence rating: 2

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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:

18.5761

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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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

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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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:

9.8696

confidence rating:

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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:

We would obtain the square root function instead of the squaring function

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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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: 2.07605

confidence rating: 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 or 3.14159

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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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: This would not be a real result so -3 wouldn't be in the first column

confidence rating: 2

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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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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:

log(2)18=x

x=ln(18)/ln(2)

x=4.16993

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

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Your solution:

log(2)12=4x

ln(12)/ln(2)=4x

x= {ln(12)/ln(2)]/4

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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Question: `q 5 * 2^x = 52

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Your solution:

2^x=52/5

log(2)(52/5)=x

ln(52/5)/ln(2)=x

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)

3x=log(9)/log(2)+4)

x= (log(9)/log(2)+4)/3

confidence rating: 2

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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):

Is that implied that this is log(10)9/log(10)2 + 4?

log does mean base-10 log.

However as we'll see, log(9) / log(2) has the same value no matter what the base, as long as both logs are taken with respect to the same base. So in this case it doesn't matter.

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

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

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Your solution:

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

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

confidence rating: 0

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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):Not real sure of myself on these....

It's not clear how you got

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

I don't think the laws of logarithms will get you from the original equation to this equation; I suspect you made an error with the algebra as well as an error with the laws of logarithms. log(a + b) is not log(a) + log(b); looks like you might have used that idea.

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

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Your solution:

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

x=log(2)/log(3)

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):

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

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Your solution:

x+1/x=log(2)/log(15)

confidence rating: 0

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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):Thats as far as I could get with this one also

&#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).

&#

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Self-critique rating:

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

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Your solution:

2^x=5^1/4

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

confidence rating: 1

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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:

Not bad overall but be sure to see my notes on a few problems that gave you trouble (and one where my solution was garbled).