Answer
a) $x=5+\log_23$
b) $x=\frac{1+\sqrt (1+4e)}{2}$
Work Step by Step
a) $2^{x-5}=3$
$x-5=\log_23$
$x=5+\log_23$
b) $\ln x+\ln (x-1)=1$, $x\gt1$
$\ln x(x-1)=1$
$x(x-1)=e$
$x^2-x=e$
$x^{2}-x-e=0$
Using Quadratic Formula
$x=\frac{-(-1)+-\sqrt ((-1)^{2} - 4(1)(-e)}{2(1)}$
Only "+" satisfies x>1
$x=\frac{1+\sqrt (1+4e)}{2}$
