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

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Answer 1 · 2018-06-20T09:37:31

#x=0 and x=ln(3)#

Let #u=e^x#. Then the equation will become #u^2-4u + 3=0#.

This equation can be solved by factoring

#(u -3)(u-1) =0#
#u = 3 or 1

Now we see that

#e^x = 3 and e^x = 1#
#x = ln3 and x= 0#

Hopefully this helps!

Answer 2 · 2018-06-20T09:41:08

#x = 0 or ln3 approx 1.09861#

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

Let #phi = e^x -> x=lnphi#

#:. phi^2 - 4phi +3=0#

#(phi-3)(phi-1)=0#

Hence, #phi = 1 or 3#

#phi =1 -> x=ln1 =0#

#phi =3 -> x= ln3 approx 1.09861#