Question

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

Answer 1

$log_{4}(x-1)=log_{2}(x-3) \\ \frac{1}{2}log_{2}(x-1)=log_{2}(x-3) \\ log_{2}(x-1)=log_{2}(x-3)^2$
$x-1=(x-3)^2=x^2-6x+9 \\ x^2-7x+10=0 \\ (x-2)(x-5)=0$
since 2-3=-1 x cannot be 2 as log₂ (-1) is not defined
so x=5 only