assignment 38

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assignment #038 ؾ{]Uӂ͛ξ College Algebra 12-11-2005

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22:50:48 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{base 3}(8) log{base 8}{(3^2) ^ log{base 3}(8) log{base 8}{(3)^2 ^ log{base 3}(8) log{base 8}(8)^2 =2

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22:51:27 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

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23:00:19 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 --> ln2/3=lna/b property #4 ln=a-b

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23:00:28 ln(2/3) = ln(2) - ln(3) = a - b.

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RESPONSE --> ok

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23:04:33 5.5.26. ln(0.5) in terms of a and b.

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RESPONSE --> ln0.5=ln1/2=ln1-ln2=b-a

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23:05:14 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

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23:29:53 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{base 3}(u^2/v)

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23:29:59 log{base 3}(u^2) log{base 3}(v) = log{base 3}(u^2 / v).

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RESPONSE --> ok

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23:37:10 5.5.68. Using a calculator express log{base1 / 2}(15)

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RESPONSE --> log{base1 / 2}(15)=log15/log.5=1.17609/-.30103=-3.907

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23:37:21 We get log{base 1/2}(15) = log(15) / log(1/2) = 1.176091259 / )-0.3010299956) = -3.906890595.

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RESPONSE --> ok

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23:49:32 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

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23:49:57 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 --> ok

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