Question

Solve the following equation:
$1-\frac{1}{2}\log (2x-1)=\frac{1}{2}\log (x-9)$

Answer 1

$1-\frac{1}{2}log(2x-1)=\frac{1}{2}log(x-9)2-log(2x-1)=log(x-9)2=log(x-9)+log(2x-1)(x-9)(2x-1)={10}^2{2x}^2-19x+9=100{2x}^2-19x-91=0{2x}^2+7x-26x-91=0x=13,\frac{-7}{2}$

But $2x-1$, $x-9$ are negative at $x=\frac{-7}{2}$ where $\log$ is undefined.

$\therefore x=13$ is only solution.