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.

037. *   37

Question:   * * *  7th edition only 5.4.14. Express the equation

2.2^3 = N

in logarithmic notation.

Your solution:

Confidence Assessment:

Given Solution:

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

Your solution:

Confidence Assessment:

Given Solution:

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

log{base 2} ( A ) = x

 in exponential notation.

Your solution:

Confidence Assessment:

Given Solution:

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

b^a = y.

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

2^x = A.

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

Your solution:

Confidence Assessment:

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.

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

Your solution:

Confidence Assessment:

Given Solution:

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

Your solution:

Confidence Assessment:

Given Solution:

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.

* * We easily solve for a by taking the -1/4 power of both sides:

(a ^ (-4))^(-1/4)  = 1/2 ^ (-1/4) so since (a^(-4))^(-1/4)) = a^(-4 * (-1/4)) = a^1 = a, and 1/2 ^(-1/4) = 2^(1/4) = sqrt( sqrt(2) ), or approximately 1.19.

a = (1/2)^-(1/4) = 2^(1/4) = sqrt( sqrt(x) ), which is approximately 1.19.

So we have

log{base sqrt(sqrt(x))} (1/2) = -4

since

(sqrt(sqrt(x)))^ (-4) = 1/2.

Note also that 1.19^(-4) is .499, very close to 1/2.

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Question:  Use transformations to graph h(x) = ln(4-x). Give domain, range, asymptotes.

Your solution:

Confidence Assessment:

Given Solution:

The graph of y = ln(x) is concave down, has a vertical asymptote at the negative y axis 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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Question: *  6.4.98 / 7th edition 5.4.102. Solve log{base 6}(36) = 5x + 3.

Your solution:

Confidence Assessment:

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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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%?

Your solution:

Confidence Assessment:

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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