# How do you solve Ln e^x = 5?

Aug 16, 2016

Real solution: $x = 5$

Complex solutions: $x = 5 + 2 k \pi i \text{ }$ for any $k \in \mathbb{Z}$

#### Explanation:

The only Real value of $x$ for which $\ln {e}^{x} = 5$ is $5$.

The function $x \mapsto {e}^{x}$ is one to one from $\left(- \infty , \infty\right)$ onto $\left(0 , \infty\right)$.

The function $x \mapsto \ln x$ is its inverse from $\left(0 , \infty\right)$ onto $\left(- \infty , \infty\right)$.

So for any $x \in \left(- \infty , \infty\right)$ we have $\ln {e}^{x} = x$.

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If we consider Complex solutions, note that ${e}^{2 \pi i} = 1$

Hence we find solutions:

$x = 5 + 2 k \pi i \text{ }$ for any integer $k$

since:

$\ln {e}^{5 + 2 k \pi i} = \ln \left({e}^{5} {e}^{2 k \pi i}\right) = \ln \left({e}^{5} \cdot {\left({e}^{2 \pi i}\right)}^{k}\right) = \ln \left({e}^{5} \cdot {1}^{k}\right) = \ln {e}^{5} = 5$