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

Apr 21, 2016

The Real solution is:

$t = \frac{1 + \ln 2}{2}$

Other Complex solutions are given by:

$t = \frac{1 + \ln 2}{2} + k \pi i$ for any $k \in \mathbb{Z}$

#### Explanation:

Given:

${e}^{2 t - 1} = 2$

Take natural logs of both sides to get:

$2 t - 1 = \ln 2$

Add $1$ to both sides to get:

$2 t = 1 + \ln 2$

Divide both sides by $2$ to find:

$t = \frac{1 + \ln 2}{2}$

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If you want all possible Complex solutions, note that:

${e}^{2 \pi i} = 1$

Hence:

${e}^{2 t - 1 - 2 k \pi i} = 2$ for any $k \in \mathbb{Z}$

So:

$2 t - 1 - 2 k \pi i = \ln 2$

Hence:

$t = \frac{1 + \ln 2 + 2 k \pi i}{2} = \frac{1 + \ln 2}{2} + k \pi i$ for any $k \in \mathbb{Z}$