SOLUTION: I need to find a solution (x,y) for the following set of numbers.... y=e^(4x) y=e^(2x)+6 I found x to be ln6/2 but i am not sure if that is right....

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I need to find a solution (x,y) for the following set of numbers.... y=e^(4x) y=e^(2x)+6 I found x to be ln6/2 but i am not sure if that is right....      Log On


   

Question 375924: I need to find a solution (x,y) for the following set of numbers....
y=e^(4x)
y=e^(2x)+6
I found x to be ln6/2 but i am not sure if that is right....
Answer by user_dude2008(1862) About Me  (Show Source):
You can put this solution on YOUR website!
y=e^(4x)
y=e^(2x)+6


e^(2x)+6=e^(4x)
(e^2x)^2-e^(2x)-6=0
z^2-z-6=0
(z-3)(z+2)=0

z=3 or z=-2

e^2x=3 or e^2x=-2


2x=ln(3)

x=ln(3)/2

Answer: x=ln(3)/2