# How do you solve 2e^x-5e^x-3=0?

##### 1 Answer
Jul 23, 2016

$2 {e}^{x} - 5 {e}^{x}$ is $- 3 {e}^{x}$, therefore your equation is equivalent to ${e}^{x} = - 1$ : no solution, because ${e}^{x} > 0$ for any real $x$.