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Math Help - Expressing In Terms of a And b

  1. #1
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    Expressing In Terms of a And b

    If log (36) = a and log (125) = b

    find log (1/12) in terms of a and b

    I tried to do things like applying the product formula to log (36), making it log (12) + log (3), and then turning log (12) into -log (1/12), but could not get any further. Any help is appreciated.
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  2. #2
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    you would set k(log36/log125)=log(1/12) find k and the rewrite the log36 and log125 with a and b (im assuming log is the natural log in this case ln??)
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  3. #3
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    I got...

    log (1/12) = -a/2+b/3-1

    could anyone confirm this? thanks!
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  4. #4
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    Hello, Shapeshift!

    Assuming that the logs are base-ten, you are right!


    If \log(36) \,=\, a\:\text{ and }\:\log(125) \,=\, b

    find \log\left(\tfrac{1}{12}\right) in terms of a\text{ and }b.
    We have: . \begin{array}{cccccccc}<br />
\log(36) \:=\:a & \Rightarrow & \log(6^2) \:=\:a & \Rightarrow & 2\log(6) \:=\:a & \Rightarrow & \log(6) \:=\:\frac{a}{2} \\ \\[-4mm]<br />
\log(125)\:=\:b & \Rightarrow & \log(5^3) \:=\:b & \Rightarrow & 3\log(5) \:=\:b & \Rightarrow & \log(5) \:=\:\frac{b}{3}<br />
\end{array}

    Therefore: . \log\left(\frac{1}{12}\right) \;=\;\log\left(\frac{5}{6\cdot10}\right) \;=\;\log(5) - \log(6) - \log(10) \;=\;\boxed{\frac{b}{3} - \frac{a}{2} - 1}

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