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

Question

308 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-12T13:37:16

Real solution: #x = 0#

Complex solutions: #x = 2npii" "# for any integer #n#

Given:

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

Note that #e^2x = (e^x)^2#

So this is effectively a quadratic equation in #e^x# ...

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

So:

#0 = (e^x)^2-2(e^x)+1 = (e^x-1)^2#

So:

#e^x = 1#

Hence real solution:

#x = 0#

Complex solutions:

#x = 2npii" "# for any integer #n#

since #e^(2npii) = (e^(2pii))^n = 1^n = 1#