Assignment 38 Mth158

#$&*

course Mth 158

12/15 9:30

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. 038. * 38 ********************************************* Question: * 6.5.24 / 6.5.18 / 7th edition 5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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) {base 3}(8) * log{base 8}(9) = 2 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 6.5.32 / 6.5.26 / 7th edition 5.5.26. ln(0.5) in terms of a and b. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln(2) = a-> ln(1) = 0 ln(.5) = ln(1/2) = ln(1) - ln(2) = 0 - a = -a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra problem / 6.5.52 / 7th edition 5.5.52. log{base 3}(u^2) - log{base 3}(v) as a single log. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * Extra Problem / 6.5.58 / 7th edition 5.5.68. Using a calculator express log{base1 / 2}(15) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: a = ln(b) -> e^a = b-> y = ln(x+c) (x+c) = e^y confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3

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Assignment 38 Mth158

#$&*

course Mth 158

12/15 9:30

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.

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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)

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

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):ok

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Self-critique Rating:3

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:3

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

ln(2) = a-> ln(1) = 0

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

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):ok

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Self-critique Rating:3

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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).

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:3

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

.............................................

Given Solution:

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

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

------------------------------------------------

Self-critique Rating:3

*********************************************

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

a = ln(b) -> e^a = b-> y = ln(x+c)

(x+c) = e^y

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

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Self-critique Rating:3

Assignment 38 Mth158

Assignment 38 Mth158

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