Question

Solve for x
$logx-\frac{1}{2}log(x-\frac{1}{2})=log(x+\frac{1}{2})-\frac{1}{2}log(x+\frac{1}{8})$

Answer 1

$x^2(x+\frac{1}{8})=(x-\frac{1}{2}){(x+\frac{1}{2})}^2{x}^3+\frac{x^2}{8}=(x-\frac{1}{2})(x^2+x+\frac{1}{4})x^3+\frac{x^2}{8}=x^3+x^2+\frac{x}{4}-\frac{x^2}{2}-\frac{x}{2}-\frac{1}{8}\frac{x^2}{8}=x^2-\frac{x^2}{2}+\frac{x}{4}-\frac{x}{2}-\frac{1}{8}\frac{x^2}{8}=\frac{x^2}{2}-\frac{x}{4}-\frac{1}{8}{x}^2=4x^2-2x-1O=3x^2=2x-1O=3x^2-3x+x-1O=3x(x-1)+1(x-1)O-(3x+1)(X-1)\Rightarrow x=1-\frac{1}{3}$