# Simplifying logarithm problem

• Jul 27th 2009, 03:06 PM
zooply
Simplifying logarithm problem
$\displaystyle 2log_{4}9 - log_{2}3$ is what i need to simplify
• Jul 27th 2009, 04:46 PM
apcalculus
Quote:

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

Note that:

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

Does this help?
• Jul 27th 2009, 04:52 PM
Prove It
Quote:

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

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.

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

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

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

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

$\displaystyle = \log_2{3}$.