# How do you solve log _2 (x-1) = log _4 (x+1)?

Dec 6, 2015

$x = 3$

#### Explanation:

First of all, you need to change the base of the logarithm: since $4 = {2}^{2}$, you have that

${\log}_{4} \left(x\right) = {\log}_{2} \frac{x}{\log} _ 2 \left(4\right) = {\log}_{2} \frac{x}{2}$

So, the equation becomes

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

Which is equivalent to

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

Since $a \log \left(x\right) = \log \left({x}^{a}\right)$, the equation becomes

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

Which is true if and only if ${\left(x - 1\right)}^{2} = x + 1$, which means

${x}^{2} - 2 x + 1 = x + 1 \setminus \to {x}^{2} - 3 x = 0$.

This equation is solved by $x = 0$ and $x = 3$, but $x = 0$ is not acceptable, because it would lead to

${\log}_{2} \left(- 1\right) = {\log}_{4} \left(1\right)$, and we can't compute logarithms of negative numbers.

As for $x = 3$, we can see that it's ok since it leads to

${\log}_{2} \left(2\right) = {\log}_{4} \left(4\right)$, which is true because both members are equal to $1$.