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Math Help - Log problems

  1. #1
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    having squared logs problems

    The question states that log3q = p

    i have to write the answer in the form of P

    -1/2logq(81q)

    thanks!
    Last edited by agehayoshina; November 30th 2012 at 10:21 AM.
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  2. #2
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    Re: Log problems

    We can write -1/2 logq⁡(81q^2 )=p as logq[(81q^2 )^-1/2], based on the laws of logs. This can be simplified to logq(1/9q) = P. This can be further simplified to logq(1/9) + logq(1/q). We know that logq(1/q) is simply -1, so we can rewrite the equation as logq(1/9) - 1 = P.Bringing 1 over to the other side gives logq(1/9) = P + 1. We can use the change of base formula to write logq(1/9) as log(1/9)/logq, giving the equation log(1/9)/logq = P + 1. Now we multiply both sides by logq to clear the equation of the denominator, resulting in log(1/9) = (P + 1)logq. We divide by (P + 1) in order to isolate q, giving logq = log(1/9)/(P + 1). Now we simply write the equation in exponential form, and find that 10^[log(1/9)/(P + 1)] = q.

    Hope this helps!
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  3. #3
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    Re: having squared logs problems

    Hello, agehayoshina!

    I hope I understand the problem.
    Here is what I think it says . . .


    \text{Given: }\,\log_3(q) = p
    \text{write in terms of }p\!:\;-\tfrac{1}{2}\log_q(81q^2)

    We have: . -\tfrac{1}{2}\log_q(81q^2) \;=\;-\log_q(81q^2)^{\frac{1}{2}} \;=\;-\log_q(9q)

    . . . . . . =\;-\big[\log_q(9) + \log_q(q)\big] \;=\;-\big[\log_q(9) + 1\big] \;=\;-\log_q(9) - 1

    . . . . . . =\;-\frac{\log_3(9)}{\log_3(q)} - 1 \;=\; -\frac{2}{p} - 1

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