$\displaystyle \log_{8}x^{\frac{3}{2}} - \log_{8}x^{\frac{1}{2}} $

Can some one please tell me how i work this out.

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- Aug 19th 2009, 05:00 PM #1

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- Aug 19th 2009, 05:09 PM #2

- Aug 19th 2009, 05:19 PM #3

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- Aug 19th 2009, 05:34 PM #4
$\displaystyle \log_8{x^{\frac{3}{2}}} - \log_8{x^{\frac{1}{2}}} = 2$

$\displaystyle \log_8\left(\frac{x^{\frac{3}{2}}}{x^{\frac{1}{2}} }\right) = 2$

$\displaystyle \log_8{x} = 2$

change to an exponential equation ...

$\displaystyle 8^2 = x$

$\displaystyle 64 = x$

in future, please post the entire question from the start.