Assignment 37 64

#$&*

course Mth 158

5/6/13 Around 8:30 AM

037. *   37 

 

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

 

2.2^3 = N

 

in logarithmic notation.

 

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Your solution:

 3 = log(base 2.2) N

 

 

confidence rating #$&*: 3

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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.

 

#### 2.2 = log(base e)(N)

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): OK

 

 

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Self-critique Rating: OK

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

 

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Your solution:

 pi = log(base x)(3)

 

 

confidence rating #$&*: 3

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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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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.

 

 

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Your solution:

 2^x = A

 

 

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

 

2^x = A.

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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:

 log{base 1/3}(9) = x

 (1/3)^x = 9

 (1/3)^x = (1/3)^-2

 x = -2

 

 

confidence rating #$&*: 2

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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.

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

 

 

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

 

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Your solution:

 x^2 - 1 > 0

 x^2 > 1

 sqrt(x^2) > sqrt(1) 

 x > 1

 or

x > -1

 x ≠ 1 and x ≠ -1 because that would equal zero and the equation must be more than zero.

 x < -1 because a negative number squared is a positive number and as long as it doesn't equal 1, it will end up greater than zero.

(-infinity, -1) U (1 , infinity) 

 

 

confidence rating #$&*: 2

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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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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).

 

 

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Your solution:

 Log{base a}(x) = y

 Log{base a}(1/2) = -4

 or

 a^(-4) = 1/2

 

 

confidence rating #$&*: 2

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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) = 16, we get

 

a = (1/2)^-(1/4) = 16.

 

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Self-critique (if necessary):

 Didn't realize I needed to solve it. I see how it was done except I'm not sure how 1/2 ^(-1/4) = 16. Could you run through that please????

@&

The given solution is in error.

a^(-b) = 1 / a^b.

So

(1/2)^(-1/4) = 1 / (1/2)^(1/4) = 2^(1/4) = 1.189, approximately.

*@

 

 

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Self-critique Rating: 2

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ln(a) is defined for a > 0.

So ln(4 - x) is defined for 4 - x > 0, i.e., for x < 4. So the domain is (-infinity, 4).

The range is all real numbers, since the equation

y = ln(4 - x)

has solution

4 - x = e^y,

or

x = 4 - e^y.

For any real y, the value of e^y is a real number, so 4 - e^y is a real number and our solution for x is also real.

*@

 

 

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

 

 

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Your solution:

 Starting with the graph of ln(x), we have the points (1/e , -1) (1, 0) and (e , 1). First, we reflect about the y-axis for ln(-x) and get the points (-1/e , -1) (-1 , 0) and (-e , 1). Then we shift the graph horizontally 4 points to the left for ln(-x + 4) and get the points (-1/e - 4 , 1) (-5 , 0) and (-e - 4 , 1).

 The vertical asymptote is now x = -4 instead of x = 0.

 The domain is 4 - x > 0, -x > -4, x < 4, so {x | x < 4}

 The range is {y | y > -4} or (-4 , infinity)

 

 

confidence rating #$&*: 2

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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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Self-critique (if necessary):

 Can you tell me what the domain and range would be????

 

 

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Self-critique Rating: 2

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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:

 log{base 6}(36) = 5x + 3

 6^(5x + 3) = 36

 6^(5x + 3) = 6^2

 5x + 3 = 2

 5x = -1

 x = -1/5

 6^(5(-1/5) + 3) = 6^2

 6^(2) = 6^2

 36 = 36

 x = -1/5

 

 

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 = 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): OK

 

 

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Self-critique Rating: OK

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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:

 F(t) = 1 - e^(-.15 t)

 0.5 = 1 - e^(-0.15 t)

 -0.5 = - e^(-0.15 t)

 0.5 = e^(-0.15 t)

 ln 0.5 = -0.15 t

 t = ln(0.5) / -0.15

 t = 4.6

 

 

confidence rating #$&*: 2

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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): OK

 

 

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Self-critique Rating: OK

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#*&!

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