# How do you solve log_2 (3x-9) - 2 = log_2 (x-4)?

##### 1 Answer
Jul 30, 2015

Get in the form log(A)=B then re-write in exponential form.

#### Explanation:

Get all logs on one side, then use quotient rule for logarithms $\log A - \log B = \log \left(\frac{A}{B}\right)$ to combine expressions:
${\log}_{2} \left(3 x - 9\right) - {\log}_{2} \left(x - 4\right) = 2$
${\log}_{2} \left(\frac{3 x - 9}{x - 4}\right)$=2

We know that ${\log}_{2} \left(4\right) = 2$, so...

$\frac{3 x - 9}{x - 4} = 4$

Multiply both sides by x-4, subtract 3x and add 16 to both sides.

x=7.

IMPORTANT: When solving log equations, check for extraneous solutions: Log of 0 or less is undefined for real numbers so make sure that substituting your value for x will not give you a 0 or negative inside the log function. In this case:

${\log}_{2} \left(12\right) - 2 = {\log}_{2} \left(3\right)$

so our solution is ok.