Answer
$x=1$
Work Step by Step
$\ln\Big(x-\dfrac{1}{2}\Big)+\ln2=2\ln x$
Combine the logarithms on the left as a product and rewrite the expression on the right as a power:
$\ln2\Big(x-\dfrac{1}{2}\Big)=\ln x^{2}$
$\ln(2x-1)=\ln x^{2}$
Since $\ln$ is one to one, the equation becomes:
$2x-1=x^{2}$
Take all terms to the right side:
$x^{2}-2x+1=0$
Solve by factoring:
$(x-1)^{2}=0$
We get one solution:
$x=1$
