# How do you solve ln4x+lnx=9?

Use the product rule for logarithms you get

$\ln 4 x + \ln x = 9$

$\ln 4 {x}^{2} = 9$

$4 {x}^{2} = {e}^{9}$

${\left(2 x\right)}^{2} = {e}^{9}$

$2 x = \sqrt{{e}^{9}}$

$x = \frac{\sqrt{{e}^{9}}}{2}$

Remark : from the defination of the logarithm we have $x > 0$