Question

Solve the following equation:
$\log x-\frac{1}{2}\log (x-\frac{1}{2})=\log (x+\frac{1}{2})-\frac{1}{2}\log (x+\frac{1}{8})$

Answer 1

$logx-\frac{1}{2}log(x-\frac{1}{2})=log(x+\frac{1}{2})-\frac{1}{2}log(x+\frac{1}{8})\Rightarrow 2logx-log(x-\frac{1}{2})=2log(x+\frac{1}{2})-log(x+\frac{1}{8})\Rightarrow log\left(\frac{x^2}{x-\frac{1}{2}}\right)=log\frac{{(x+\frac{1}{2})}^2}{(x+\frac{1}{8})}\Rightarrow \frac{{2\left(x\right)}^2}{2x-1}=\frac{2(2x+1{)}^2}{8x+1}\Rightarrow x^2(8x+1)=(2x-1)({2x+1)}^2\Rightarrow {8x}^3+x^2=({4x}^2-1\left)\right(2x+1)\Rightarrow {8x}^3+x^2={8x}^3+{4x}^2-2x-1\Rightarrow {3x}^2-2x-1=0\Rightarrow {3x}^2-3x+x-1=0\Rightarrow (3x+1\left)\right(x-1)=0$

Therefore, $x=\frac{-1}{3}$ or $1.$

But $x\ne \frac{-1}{3}(\because \log x,\log (x-\frac{1}{2})$ are undefined at $x=\frac{-1}{3})$

Therefore, $x=1$ is the only solution.