Assign 37

course mth 158

As I was reading through the chapter, I understood each section (with maybe some difficulty in the word problems) but I felt that I had a really good understanding of the chapter until I tried to solve the problems here. I don't know how to improve my skills on solving problems, but I keep trying.

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.

036. * 36

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Question: * 6.3.40 / 7th edition 5.3.38. Explain how you used transformations to graph f(x) = 2^(x+2)

I started with the graph of y = 2^x which has the points (-1, ½), (0, 1), (1, 2). Then y = 2^(x+2) shifts the graph 2 units to the right which now has the points (1, ½), (2, 1), (3, 2). The graph gets closer to the x axis as x moves towards –infinity, and increases at an increasing rate as x moves towards infinity.

The function y = 2 ^ ( x+2) is a transformation of the basic function y = 2^x, with x replaced by x - 2. So the graph moves 2 units in the x direction.

The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2). The graph of y = 2^(x-2), being shifted in only the x direction, is also asymptotic to the negative x axis and passes through the points (0 + 2, 1) = (2, 1) and (1 + 2, 2) = (3, 2). The graph also increases at a rapidly increasing rate.

All the points of the graph of y = 2^(x-2) lie 2 units to the right of points on the graph of y = 2^x.

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Self-critique (if necessary):

Why would you replace x with x - 2 in this problem? And when you replace it, why does the graph move 2 units right? At first when I tried to solve the problem, I followed one of the examples in the book where you create a table with different points such as:

x = -3, -2, -1, 0, 1, 2, 3 and plugged these numbers into 2^(x+2) to get y = ½, 1, 2, 4, 8, 16, 32

But the graph didn’t look like y = 2^ x (pictured in the book on page 428) so I did the graph based on that.

Why does the book teach using tables, but when you do, the graph is not the same as the given answer?

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Self-critique Rating: 3

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Question: * Extra Problem / 7th edition 5.3.42. Transformations to graph f(x) = 1 – 3 * 2^x

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Your solution:

Problem 5.3.42 is a different problem. It is f(x) = 2x^2 + 16/x (probably because I have the 8th edition)

I tried to solve both of these, but just get stuck. The book says to follow the steps outlines on page 355, but they don’t make sense to me on either of these problems.

confidence rating: 0

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Given Solution:

* * 1 - 3 * 2^x is obtained from y = 2^x by first vertically stretching the graph by factor -3, then by shifting this graph 1 unit in the vertical direction.

The graph of y = 2^x is asymptotic to the negative x axis, increases at a rapidly increasing rate and passes through ( 0, 1 ) and (1, 2).

-3 * 2^x will approach zero as 2^x approaches zero, so the graph of y = -3 * 2^x will remain asymptotic to the negative x axis. The points (0, 1) and (1, 2) will be transformed to (0, -3* 1) = (0, -3) and (1, -3 * 2) = (1, -6). So the graph will decrease from its asymptote just below the x axis through the points (0, -3) and (1, -6), decreasing at a rapidly decreasing rate.

To get the graph of 1 - 3 * 2^x the graph of -3 * 2^x will be vertically shifted 1 unit. This will raise the horizontal asymptote from the x axis (which is the line y = 0) to the line y = 1, and will also raise every other point by 1 unit. The points (0, -3) and (1, -6) will be transformed to (0, -3 + 1) = (0, -2) and (1, -6 + 5) = (1, -5), decreasing from its asymptote just below the line y = 1 through the points (0, -2) and (1, -5), decreasing at a rapidly decreasing rate.

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Self-critique (if necessary):

I still really don’t understand this one.

Check out the original q_a_ document for the additional explanations I've added. The explanations involve tables and graphs, and wouldn't represent properly in this document. Just click on

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ca/ca_query36.htm

and your browser should take you to that page.

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Self-critique Rating: 0

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Question: * Extra Problem / 7th edition 5.3.60 Solve (1/2)^(1-x) = 4.

(My 5.3.60 problem from 8th edition is a word problem)

The only example in the book that looks like this problem and how it might be solved is this:

½^1-x = 4

ln4 = 1 – x

-x = ln4/1

-x approx = 1.39 divided by -1

x approx = -1.39

but I am not even sure I done that right either.

(1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2.

If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3.

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Self-critique (if necessary):

I have really messed up on these problems. I have spent the better part of this weekend reading and doing the problems in chapter 6. I thought I had a good grasp on the concepts of this chapter. It was actually fun doing those problems, but after trying to do the ones here I am not so sure I have a grasp on this.

The given solution to this problem just doesn’t make any sense to me. I can’t follow along with how the answer was obtained.

It's worth looking at the revised document for this problem. The formatting in the following is better on the original document:

The reliable way to solve a problem of this nature is to recognize that the variable x occurs in an exponent, so we can 'get at it' by taking logs of both sides:

(1/2)^(1-x) = 4. Taking the natural log of both sides, using the laws of logarithms, we get

ln( (1/2)^(1 - x) = ln(4)) so that

(1 - x) ln(1/2) = ln(4) and

(1 - x) = ln(4) / ln(1/2). A calculator will reveal that ln(4) / ln(1/2) = -2, so that

1 - x = -2. We easily solve for x, obtaining

x = 1.

It is also possible to reason this problem out directly, and in this case our reasoning leads us to an exact solution:

We first recognize one fact:

4 is an integer power of 2, and 1/2 is an integer power of 2, so 4 must also be an integer power of 1/2.

Since 4 = 2^2, and since 1/2 = 2^-1, we can recognize that 4 = 2^-2, and reason as follows:

(1/2)^(-2) = 1 / (1/2)^2 = 1 / (1/4) = 4, so we know that the exponennt 1 - x must be -2

If 1 - x = -2, it follows that -x = -2 - 1 = -3 and x = 3.

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Self-critique Rating: 1

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Question: * 6.3.98 / 7th edition 5.3.78 A(n) = A0 e^(-.35 n), area of wound after n days. What is the area after 3 days and what is the area after 10 days? Init area is 100 mm^2.

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Your solution:

If the initial area of the wound is 100 mm^2 then to solve the problem you would

A(n) = A0 e^(-0.35n) by substituting 3 for n you would have

100mm^2 * e^(-0.35(3)) approx = 34.994mm^2

and for 10 days

100mm^2 * e^(-0.35(10)) approx = 3.020mm^2

confidence rating: 2

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Given Solution:

* * If the initial area is 100 mm^2, then when n = 0 we have A(0) = A0 e^(-.35 * 0) = 100 mm^2.

Since e^0 = 1 this tells us that A0 = 100 mm^2.

So the function is A(n) = 100 mm^2 * e^(-.35 n).

To get the area after 3 days we evaluate the function for n = 3, obtainind A(3) = 100 mm^2 * e^(-.35 * 3) = 100 mm^3 * .35 = 35 mm^2 approx..

After to days we find that the area is A(10) = 100 mm^2 * e^(-.35 * 10) 100 mm63 * .0302 = 3.02 mm^2 approx..

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Self-critique (if necessary):

I don’t think there are any discrepancies. I can’t believe I solved this word problem without help!!

Very well done.

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Self-critique Rating: 3

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Question: * 6.3.104 / 7th edition 5.3.84. Poisson probability 4^x e^-4 / x!, probability that x people will arrive in the next minute. What is the probability that 5 will arrive in the next minute, and what is the probability that 8 will arrive in the next minute?

The given solution was just beneath the question, but I had solved this earlier on paper before I see what the solution says, I will give my answer. I am not so confident that I solved this one like the previous one but here goes:

P(x) = 4^x e^-4 / x!

The probability that 5 people will arrive in the next minute would be approximately 0.15629 and for 8 people the probability is approx 0.02977

I just put x=4 4^4 * e^-4/4!

Then x=8 4^8 * e^-4/8!

The probability that 5 will arrive in the next minute is P(5) = 4^5 * e^-4 / (5 !) = 1024 * .0183 / (5 * 4 * 3 * 2 * 1) = 1024 * .0183 / 120 = .152 approx..

The probability that 8 will arrive in the next minute is P(8) = 4^8 * e^-4 / (8 !) = .030 approx..

037. * 37

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Your solution:

I didn’t get the second part right. I’m not sure why the given solution is e^-8 why would -4 which is a constant, change to -8 here?

There's a typo; should read 4^8 * e^(-4) / 8!. I've corrected this above.

confidence rating: 3

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Given Solution:

* * * 7th edition only 5.4.14. 2.2^3 = N. Express in logarithmic notation.

Read as 3 is the logarithm at base 2.2 of N expressed in notation would be

3 = log{base 2.2}N

b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 2.2, x = 3 and y = N.

So we write lob{base b}(y) = x as

log{base 2.2}(N) = 3.

8th edition only 6.4.14. e^2.2 = N. Express in logarithmic notation.

b^x = y is expressed in logarithmic notation as log{base b}(y) = x. In this case b = 3, x = 2.2 and y = N.

So we write lob{base b}(y) = x as

log{base e}(N) = 2.2.

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Self-critique (if necessary):

I think I wrote it backwards. I have a good basic understanding of the logarithmic function so far. (I hope)

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Self-critique Rating: 3

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Question: * extra problem / 7th edition 5.4.18. x^pi = 3. Express in logarithmic notation.

pi is the logarithm at base x of 3

log{base x}(3) = pi

b^a = y is expressed in logarithmic notation as log{base b}(y) = a. In this case b = x, a = pi and y = 3.

So we write lob{base b}(y) = a as

log{base x}(3) = pi.

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Self-critique (if necessary):

There are no discrepancies

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Self-critique Rating: 3

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Question: * extra problem / 7th edition 5.4.26.. log{base 2}?? = x. Express in exponential notation.

x is the logarithm of base 2 of y or

2^x = y

log{base b}(y) = a is expressed in exponential notation as

b^a = y.

In this case b = 2, a = x and y = ?? so the expression b^a = y is written as

2^x = ??.

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Self-critique (if necessary):

I’m not sure if I should have put the y in my equation.

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Self-critique Rating: 2

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Question: * 6.4.30 / 7th edition 5.4.36. Exact value of log{base 1/3}(9)

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Your solution:

To solve log{base 1/3}(9) you set y = log{base 1/3}(9)

y = log{base 1/3}(9)

1/3^y = 9

1/3^y = 1/3^-2 we know that 3^2 = 9 so the reciprocal of 2 which is -2 should work here

y = -2

Therefore, log{base 1/3}(9) = -2

confidence rating: 3

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Given Solution:

* * log{base b}(y) = a is expressed in exponential notation as

b^a = y.

In this case b = 1/3, a is what we are trying to find and y = 9 so the expression b^a = y is written as

(1/3)^a = 9.

Noting that 9 is an integer power of 3 we expect that a will be an integer power of 1/3. Since 9 = 3^2 we might try (1/3)^2 = 9, but this doesn't work since (1/3)^2 = 1/9, not 9. We can correct this by using the -2 power, which is the reciprocal of the +2 power: (1/3)^-2 = 1/ ( 1/3)^2 = 1 / (1/9) = 9.

So log

base 1/3}(9) = -2.

. What is the domain of G(x) = log{base 1 / 2}(x^2-1)

For any positive value of b the domain of log{base b}(x) consists of all positive real numbers. It follows that log{base 1/2}(x^2-1) consists of all real numbers for which x^2 - 1 > 0.

We solve x^2 - 1 > 0 for x by first finding the values of x for which x^2 - 1 = 0. We can factor the expression to get (x-1)(x+1) = 0, which is so if x-1 = 0 or if x + 1 = 0. Our solutions are therefore x = 1 and x = -1.

It follows that x^2 - 1 is either positive or negative on each of the intervals (-infinity, -1), (-1, 1) and (1, infinity). We can determine which by substituting any value from each interval into x^2 - 1.

On the interval (-1, 1) we can just choose x = 0, which substituted into x^2 - 1 gives us -1. We conclude that x^2 - 1 < 0 on this interval.

On the interval (-infinity, -1) we could substitute x = -2, giving us x^2 - 1 = 4 - 1 = 3, which is > 0. So x^2 - 1 > 0 on (-infinity, -1).

Substituting x = 2 to test the interval (1, infinity) we obtain the same result so that x^2 - 1 > 0 on (1, infinity).

We conclude that the domain of this function is (-infinity, -1) U (1, infinity).

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Self-critique (if necessary):

I don’t see any discrepancies, but I’m not sure where this came from:

. What is the domain of G(x) = log{base 1 / 2}(x^2-1)

That is another problem, followed by a solution. This wasn't clearly marked in the original.

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Self-critique Rating: 3

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Question: * 6.4.58 / 7th edition 5.4.62. a such that graph of log{base a}(x) contains (1/2, -4).

log{base a}(x) = y if a^y = x. The point (1/2, -4) will lie on the graph if y = -4 when x = 1/2, so we are looking for a value of a such that a ^ (-4) = 1/2.

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Your solution:

I don’t know how to solve this one.

confidence rating: 0

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Given Solution:

* * We easily solve for a by taking the -1/4 power of both sides, obtaining a = (1/2)^-(1/4) = 16.

. Transformations to graph h(x) = ln(4-x). Given domain, range, asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = ""

The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axilineCount = lineCount + 1: bLine$(lineCount) = s and passes through the points (1, 0) and (e, 1) (the latter is approximately (2.71828, 1) ).

The graph of y = ln(x - 4), where x is replaced by x - 4, is shifted +4 units in the x direction so the vertical asymptote shifts 4 units right to the line x = 4. The points (1, 0) and (e, 1) shift to (1+4, 0) = (5, 0) and (e + 4, 1) (the latter being approximately (3.71828, 1).

Since x - 4 = - (4 - x), the graph of y = ln(4 - x) is 'flipped' about the y axis relative to ln(x-4), so it the vertical asymptote becomes x = -4 and the graph passes through the points (-5, 0) and (-e-4, 1). The graph will still be concave down.

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Self-critique (if necessary):

I couldn’t follow along with the solution on this one. It was a little confusing. Also, I think some of your code is showing up in the solution: it may be missing a ( ) or “ ”

asymptotes.lineCount = lineCount + 1: bLine$(lineCount) = ""

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Self-critique Rating:

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Question: * 6.4.98 / 7th edition 5.4.102. Solve log{base 6}(36) = 5x + 3.

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Your solution:

This is one that I tried all evening to solve. It looks like it is

5x+3 is the logarithm of base 6 at 36 so would you set it up like this:

6^5x+3 = 36 we know that 6^2 = 36 so 5x+3 has to equal 2?

5x+3 = 2

5x = 2-3

5x = 1

x = 1/5

6^5(1/5) + 3 = 1296 so that’s not right either.

confidence rating: 0

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Given Solution:

* * log{base b}(y) = a is expressed in exponential notation as

b^a = y.

In this case b = 6, a is what we are trying to find and y = 5x + 3 so the expression b^a = y is written as

6^(5x+3) = 36.

We know that 6^2 = 36, so (5x + 3) = 2. We easily solve this equation to get x = -1 / 5.

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Self-critique (if necessary):

I was on the right track. But how did you get -1/5?

5x + 3 = 2, so 5x = -1 and x = -1/5.

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Self-critique Rating: 3

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Question: * 6.4.122 / 7th edition 6.3.102. F(t) = 1 – e^(-.15 t) prob of car arriving within t minute of 5:00. How long does it take for the probability to reach 50%? How long to reach 80%?

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Your solution:

50% = .5 and 80% = .8

Would you set 1-e^(-.15 t) = .5 ? I really am not sure how to solve this.

confidence rating: 0

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Given Solution:

* * The probability is F(t), so F(t) = 50% = .5 when 1 - e^(-.15 t) = .5.

We subtract 1 from both sides the multiply by -1 to rearrange this equation to the form

e^(-.15 t) = .5.

Taking the natural log of both sides we get

ln( e^(-.15 t) ) = ln(.5), which tells us that

-.15 t = ln(.5) so that

t = -ln(.5)/.15 = 4.6, approx..

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Self-critique (if necessary):

I tried doing what the given solution suggests and subtracted 1 from both sides, but mine was

-e^(-.5t) = -1/2 so I must have done something wrong. Then, when I tried to do it step by step with the given solution, I entered ln(e^-.15) and came up with -0.15 and ln(.5) = -0.69 so how does that equal. I am not clear on how to solve this equation.

ln(.5) is about -.69, so

-ln(.5) / .15 = -.69 / .15 = 4.6.

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Self-critique Rating: 1

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

Note also that the qa's for assts 36 and 37 were originally on one file; I've corrected that and I believe the two documents are now properly linked on the Assignments page.