$ln$

$ln(0.5)=ln(\frac{1}{2})$

$=ln(2^{-1})$

$=-ln(2)$

So the solution is $\boxed{2}$ .

$\ln\left(0.5\right) = -\ln\left(x\right)\\ \ln\left(2^{-1}\right) = \ln\left(x^{-1}\right)\\ x = \color{#69047E}{\boxed{2}}$