# How do you solve log_4(3x-2)-log_4(4x+1)=2?

Jul 9, 2016

$x = - \frac{18}{61}$
This answer can be verified by substituting in the original equation

#### Explanation:

The first thing we need to do is to write 2 as a ${\log}_{4}$ term as well.
With logs, the terms must either be all as logs, or all as numbers, not a combination.

$2 = {\log}_{4} 16 , \text{ because } {4}^{2} = 16$

${\log}_{4} \left(3 x - 2\right) - {\log}_{4} \left(4 x + 1\right) = {\log}_{4} 16$

If we are subtracting the logs, we must have been dividing the numbers. It is possible to condense two log terms into one using the log laws. $\log A - \log B = \log \left(\frac{A}{B}\right)$

${\log}_{4} \left(\frac{3 x - 2}{4 x + 1}\right) = {\log}_{4} 16$

Now because there is one term on each side, the terms which we are finding logs of, must be equal to each other. Drop the logs.

$\frac{3 x - 2}{4 x + 1} = 16 \text{ solve as usual}$

$16 \left(4 x + 1\right) = 3 x - 2$

$64 x + 16 = 3 x - 2$

$61 x = - 18$

$x = - \frac{18}{61}$