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

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Answer 1 · 2016-04-21T22:55:46

The Real solution is:

$t = (1 + ln 2)/2$

Given:

$e^(2t-1) = 2$

Take natural logs of both sides to get:

$2t-1 = ln 2$

Add $1$ to both sides to get:

$2t = 1 + ln 2$

Divide both sides by $2$ to find:

$t = (1+ln 2)/2$

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If you want all possible Complex solutions, note that:

$e^(2pi i) = 1$

Hence:

$e^(2t-1-2k pi i) = 2$ for any $k in ZZ$

So:

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

Hence:

$t = (1 + ln 2 + 2k pi i)/2 = (1 + ln 2)/2 + k pi i$ for any $k in ZZ$

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