Assignment 38 65

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course Mth 158

5/6/13 Around 1:30 PM

038. *   38 

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Question: *  6.5.24 / 6.5.18 / 7th edition 5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9).

 

 

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Your solution:

 log{base 3}{(8) * log{base 8}(9)

 (log(8) / log(3) * (log(9) / log(8)

 log(9) / log(3)

 log{base 3}(9)

 3^y = 9

 3^y = 3^2

 y = 2

 

 

confidence rating #$&*: 3

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Given Solution:

* * log{base 3}(8) * log{base 8}(9) = log 8 / log 3 * log 9 / log 8 = log 9 / log 3 = log{base 3}(9).

 

log{base 3}(9) is the power to which 3 must be raised to get 9, and is therefore equal to 2.

 

Thus log{base 3}(8) * log{base 8}(9) = 2.

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: *  6.5.30 / 6.5.24 / 7th edition 5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b?

 

 

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Your solution:

 ln(2/3) = ln(2) - ln(3)

 a - b

 

 

confidence rating #$&*: 3

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Given Solution:

* * ln(2/3) = ln(2) - ln(3) = a - b.

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: *  6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b.

 

 

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Your solution:

 I am not sure.

 

 

confidence rating #$&*: 1

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Given Solution:

* * Since ln(2) = a, and since ln(1) = 0, we have

 

ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a.

 

 

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Self-critique (if necessary):

 I didn't think of turning 0.5 into 1/2. I totally see how it works now.

 

 

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Self-critique Rating: 3

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Question: *  Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log.

 

 

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Your solution:

 log{base 3}(u^2) - log{base 3}(v)

 log{base 3}(u^2 / v)

 

 

confidence rating #$&*: 3

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Given Solution:

* * log{base 3}(u^2) - log{base 3}(v) = log{base 3}(u^2 / v).

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: *  Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15)

 

 

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Your solution:

 log{base1 / 2}(15) = -3.88 approximate.

 

 

confidence rating #$&*: 3

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Given Solution:

* * We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595.

 

 

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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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Question: *  6.5.88 / 6.5.82 / 7th edition 5.5.80. Express y as a function of x if ln y = ln(x + C).

 

 

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Your solution:

 y = x + C

 

 

confidence rating #$&*: 2

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Given Solution:

* * a = ln(b) means e^a = b, so y = ln(x+c) is translated to exponential form as

 

(x+c) = e^y.

 

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Self-critique (if necessary):

 I'm not sure I understand the given solution. Could you explain please????

 

 

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Self-critique Rating: 1

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It appears that the given solution was for the equation

y = ln(x + C).

Your solution was correct for the given equation.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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