Simplifying logarithm problem

**zooply** · July 27th 2009, 03:06 PM

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

**apcalculus** · July 27th 2009, 04:46 PM

Originally Posted by zooply

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

Note that:

$log_{2}3 = \frac{log_{10} 3}{log_{10} 2} = 2 \frac{log_{10} 3}{2 log_{10} 2} = 2 \frac{log_{10} 3}{log_{10} 4} =2 log_{4} 3 = log_{4} 9$

Does this help?

**Prove It** · July 27th 2009, 04:52 PM

Originally Posted by zooply

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

You need to use the change of base rule

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

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

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

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

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

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

$= \log_2{3}$ .

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