# How do you solve e^x>1.6?

Feb 7, 2017

$x > \ln 1.6$, with $\ln 1.6 \approx 0.470003629$.

#### Explanation:

Since $\ln x$ is a constantly increasing function, for any $a > b$ it holds that $\ln a > \ln b$. Applying this to the relationship:

$\ln {e}^{x} > \ln 1.6$, and we know that $\ln {e}^{x} = x$

If even further confirmation is required, from logarithm properties:

$\ln {e}^{x} = x \ln e = x \cdot 1 = x$