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....
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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) (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
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