How do you solve ${log}_{2} (x - 1) = {log}_{4} (x + 1)$ $log_2 (x-1) = log_4 (x+1)$ ?

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Answer 1 · 2016-01-12T23:28:14

$S = {0, 3}$ $S = {0,3}$

Elevate four to both sides

$4^{{log}_{2} (x - 1)} = 4^{{log}_{4} (x + 1)}$ $4^(log_2(x-1)) = 4^(log_4(x+1))$

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

$4^{{log}_{2} (x - 1)} = x + 1$ $4^(log_2(x-1)) = x+1$

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

$2^{2 {log}_{2} (x - 1)} = x + 1$ $2^(2log_2(x-1)) = x+1$

So we have

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

Expanding

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

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

How do you solve log2(x−1)=log4(x+1)log_2 (x-1) = log_4 (x+1)?