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

##### 1 Answer

Apr 21, 2016

The Real solution is:

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

Other Complex solutions are given by:

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

#### Explanation:

Given:

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

Take natural logs of both sides to get:

#2t-1 = ln 2#

Add

#2t = 1 + ln 2#

Divide both sides by

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

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#