# How do you solve e^(4x-1)=(e^2)^x ?

$x = \frac{1}{2}$
Well it is ${e}^{4 x - 1} = {e}^{2 x} \implies {e}^{4 x - 2 x - 1} = 1 \implies {e}^{2 x - 1} = 1 \implies 2 x - 1 = 0 \implies x = \frac{1}{2}$