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

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Answer 1 · 2016-12-31T11:09:12

The answer is #x={ln4}#

Let #e^x=y#

Then the equation

#e^(2x)-3e^x-4=0#

becomes

#y^2-3y-4=0#

We solve this as a quadratic equation

We calculate the discriminant

#Delta=b^2-4ac=9-4*1*-4=9+16=25#

#Delta>0#, we have 2 real solutions

#y_1=(-b+sqrtDelta)/2a=(3+5)/2=4#

#y_1=(-b-sqrtDelta)/2a=(3-5)/2=-1#

Therefore

#e^x=4#, #=>#, #x=ln4#

#e^x=-1#, #=>#, no solution as #AA x in RR, e^x>0#