Log needs a log

Solving (x - 1) = (x - 3)^2 we get X = 2, 5. As log is valid for positive real only x = 5 is a solution.

$log_{4} (x - 1)$ = $log_{2} (x - 3)$

$2^{x - 1} = 4^{x - 3}$

$2^{x - 1} = 2^{2(x - 3)}$

$log_{2} 2^{x - 1}$ = $log_{2} 2^{2(x - 3)}$

x - 1 = 2( x - 3)

x - 1 = 2x - 6

x = 6 - 1

x = 5

Only one solution is obtained