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

1 Answer
Apr 21, 2016

Answer:

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 #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#