Answer
a) $x=5+\log_23$
b) $x=\frac{1}{2}(1+\sqrt{4e+1})$
Work Step by Step
a) $2^{x-5}=3$
$x-5=\log_23$
$x=5+\log_23$
b) $\ln x+\ln (x-1)=1$, $x\gt1$
$\ln x(x-1)=1$
$x(x-1)=e$
$x^2-x=e$
$x^2-x+\frac{1}{4}=e+\frac{1}{4}$
$(x-\frac{1}{2})^2=e+\frac{1}{4}$
$x-\frac{1}{2}=\pm\sqrt{e+\frac{1}{4}}$
$x=\frac{1}{2}\pm\sqrt{\frac{1}{4}(4e+1)}\gt0$
$\therefore x=\frac{1}{2}+\frac{1}{2}\sqrt{4e+1}$
$=\frac{1}{2}(1+\sqrt{4e+1})$
