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Question

#e^(4x) + 4e^(2x) - 21 = 0#

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Answer 1 · 2017-04-13T12:02:50

#x=1.0986#

In #e^(4x) + 4e^(2x) - 21 = 0#, let #e^(2x)=u#

then the equation becomes

#u^2+4u-21=0#

or #u^2+7u-3u-21=0#

or #u(u+7)-3(u+7)=0#

or #(u-3)(u+7)=0#

bur as #u=e^x#, #u+7!=0# and dividing by #u+7#, we get

#u-3=0# or #e^x=3# i.e. #x=ln3=1.0986#