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

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Answer 1 · 2015-12-06T11:00:14

$x=3$

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

$log_4(x) = log_2(x)/log_2(4) = log_2(x)/2$

So, the equation becomes

$log_2(x-1) = log_2(x+1)/2$

Which is equivalent to

$2log_2(x-1) = log_2(x+1)$

Since $alog(x)=log(x^a)$ , the equation becomes

$log_2((x-1)^2) = log_2(x+1)$

Which is true if and only if $(x-1)^2=x+1$ , which means

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

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

$log_2(-1)=log_4(1)$ , and we can't compute logarithms of negative numbers.

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

$log_2(2) = log_4(4)$ , which is true because both members are equal to $1$ .

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