How do you solve #e^(4x)-3e^(2x)-18=0#?

Question

5082 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-05T16:38:17

#x=1/2ln6#

To solve #e^(4x)-3e^(2x)-18=0#, let #e^(2x)=u#

Then the above equation becomes

#u^2-3u-18=0# and splitting middle term we get

#u^2-6u+3u-18=0#

or #u(u-6)+3(u-6)=0# or #(u+3)(u-6)=0#

substituting #u=e^(2x)#, we get

#(e^(2x)+3)(e^(2x)-6)=0#

As #e^(2x)# is always positive

#(e^(2x)+3)!=0# and hence dividing above by #(e^(2x)+3)#,

#(e^(2x)-6)=0#

i.e. #e^(2x)=6#

or #2x=ln6# and hence #x=1/2ln6#