$\displaystyle 2log_{4}9 - log_{2}3$ is what i need to simplify

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- Jul 27th 2009, 03:06 PMzooplySimplifying logarithm problem
$\displaystyle 2log_{4}9 - log_{2}3$ is what i need to simplify

- Jul 27th 2009, 04:46 PMapcalculus
- Jul 27th 2009, 04:52 PMProve It
You need to use the change of base rule

$\displaystyle \log_b{x} = \frac{\log_k{x}}{\log_k{b}}$.

I always use natural logs, but you can choose whatever base you like.

$\displaystyle 2\log_4{9} - \log_2{3} = \frac{2\ln{9}}{\ln{4}} - \frac{\ln{3}}{\ln{2}}$

$\displaystyle = \frac{4\ln{3}}{2\ln{2}} - \frac{\ln{3}}{\ln{2}}$

$\displaystyle = \frac{2\ln{3}}{\ln{2}} - \frac{\ln{3}}{\ln{2}}$

$\displaystyle = \frac{\ln{3}}{\ln{2}}$

$\displaystyle = \log_2{3}$.