Assignment 38

course MTH 158

š˜ÃD͈qµól°ù’öå­ì³Þ±¼ôassignment #038

038. Query 38

College Algebra

08-05-2008

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13:46:32

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) =

log{base 8}(9)^(log{base3}(8)) =

log{base 8}(3^2)^(log{base3}(8)) =

log{base 8}(3)^log{base 3}(8)^2 =

log{base 8}(8^2) =

2

confidence assessment: 3

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13:46:37

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 -->

ok

self critique assessment: 3

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13:46:56

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 -->

a - b

confidence assessment: 3

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13:47:00

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

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RESPONSE -->

ok

self critique assessment: 3

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13:47:13

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

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RESPONSE -->

ln(0.5) = ln(1/2)

ln(1) - ln(2)

0 - a = -a

confidence assessment: 2

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13:47:17

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 -->

ok

self critique assessment: 3

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13:47:30

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/v)

confidence assessment: 3

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13:47:37

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

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RESPONSE -->

ok

self critique assessment: 3

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13:49:03

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)) =

(log 15) / (-log 2) = approx

1.17609 / -.30103 = approx

-3.907

confidence assessment: 2

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13:49:10

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

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RESPONSE -->

ok

self critique assessment: 3

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13:54:35

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

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RESPONSE -->

y = x + c

confidence assessment: 2

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13:55:43

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 -->

I don't understand where the e^y comes in, that wasn't in the lesson...

self critique assessment: 1

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The natural log function, abbreviated ln as opposed to log, is the base-e log.

ln(x + C) means log{base e}(x + C), so that

y = ln(x+C) means

y = log{base e} (x + C). This translates to

e^y = x + C.

&#This looks good. See my notes. Let me know if you have any questions. &#