# How do you solve for x in 2e^(x-2)=e^x + 7?

Jun 14, 2016

$x = \ln 7 - \ln \left(2 {e}^{- 2} - 1\right)$

#### Explanation:

$2 {e}^{x - 2} = {e}^{x} + 7$

or $2 {e}^{x} / {e}^{2} = {e}^{x} + 7$

or $2 {e}^{x} / {e}^{2} - {e}^{x} = 7$

or ${e}^{x} \left(2 {e}^{- 2} - 1\right) = 7$

or e^x=7/((2e^(-2)-1)

or $x = \ln \left(\frac{7}{\left(2 {e}^{- 2} - 1\right)}\right)$

or $x = \ln 7 - \ln \left(2 {e}^{- 2} - 1\right)$