# How do you solve log_2 (x-1) = log_4 (x+1)?

##### 1 Answer
Jan 12, 2016

$S = \left\{0 , 3\right\}$

#### Explanation:

Elevate four to both sides

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

Since ${a}^{{\log}_{a} \left(b\right)} = b$ we can simplify the RHS

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

But since $4 = {2}^{2}$

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

So we have

${\left(x - 1\right)}^{2} = x + 1$

Expanding

${x}^{2} - 2 x + 1 = x + 1$
${x}^{2} - 2 x - x + 1 - 1 = 0$
${x}^{2} - 3 x = 0$

The roots are $x = 0$ and $x = 3$