Assign 38

course Mth 158

Mr. Smith,I was wondering if there is more q_a querys after this one, 38. My computer is telling me that it was out of range when I tried to work 39 and 40. Also, about chapter 8 test...Is there one? I have also finished test R, 1, and 2.

Test 8 isn't being used this term, so when you're done with the test in exponential function, you're finished. I don't think there are any queries beyond 38.

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assignment #038

038. Query 38

College Algebra

04-28-2007

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19:28:54

5.5.18. Exact value of log{base 3}{(8) * log{base 8}(9).

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RESPONSE -->

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

using the change-of-base formula

=log 8 / log 3 * log 9 / log 8

=log 9 / log 3

=log{base 3}(9) = 2

confidence assessment: 2

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19:29:16

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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RESPONSE -->

Yes! I got that one right!

self critique assessment: 2

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19:33:50

5.5.24. ln(2) = a, ln(3) = b. What is ln(2/3) in terms of a and b?

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RESPONSE -->

ln(2) = a, ln(3) = b

ln2/3

ln2 - ln3 = a - b

confidence assessment: 2

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19:40:43

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

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RESPONSE -->

whooah!

self critique assessment: 2

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19:50:51

5.5.26. ln(0.5) in terms of a and b.

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RESPONSE -->

ln(0.5)

ln(1/2)

ln1 - ln2 ln(2)=a ln(1)=0

0 - a

- a

confidence assessment: 2

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19:50:54

5.5.26. ln(0.5) in terms of a and b.

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RESPONSE -->

confidence assessment:

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19:51:15

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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RESPONSE -->

okay!

self critique assessment: 2

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19:52:15

5.5.52. log{base 3}(u^2) ·log{base 3}(v) as a single log.

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RESPONSE -->

log{base 3}(u^2) ·log{base 3}(v)

log{base3}(u^2/v)

confidence assessment: 2

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19:52:27

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

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RESPONSE -->

whoooah!

self critique assessment: 2

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19:54:14

5.5.68. Using a calculator express log{base1 / 2}(15)

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RESPONSE -->

log{base1 / 2}(15)

log(15)/log(1/2)= approximately

1.1761/- 0.3010= approximately

- 3.9069

confidence assessment: 2

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19:54:25

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

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RESPONSE -->

okay!

self critique assessment: 2

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19:56:44

5.5.80. Express y as a function of x if ln y = ln(x + C).

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RESPONSE -->

ln y = ln(x + C)

y=x+C

confidence assessment: 2

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19:57:27

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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RESPONSE -->

hmmm...okay!

self critique assessment: 2

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Your work looks good. Let me know if you have questions.