Math help

e4x - 5 * e2x = -6 

My answer will be similar I think.

\(e^{4x}-5e^{2x}=6\\\\ (e^{2x})^2-5e^{2x}-6=0\\\\ Sub\;\;y=e^{2x}\\\\ y^2-5y-6=0\\\\ (y-6)(y+1)=0\\\\ y=6\;\;or\;\;y=-1\\\\ e^x=6\;\;or\;\;e^x=-1\\\\ e^x=6\;\;or\;\;no\; solutions\\\\ lne^x=ln6\\\ x=ln6 \)

Answering guest, I think you made a little mistake when you completed the square :)

Solve for x over the real numbers:
e^(4 x)-5 e^(2 x) = -6

Simplify and substitute y = e^(2 x):
e^(4 x)-5 e^(2 x)  =  (e^(2 x))^2-5 e^(2 x)  =  y^2-5 y  =  -6:
y^2-5 y = -6

Add 25/4 to both sides:
y^2-5 y+25/4 = 1/4

Write the left hand side as a square:
(y-5/2)^2 = 1/4

Take the square root of both sides:
y-5/2 = 1/2 or y-5/2 = -1/2

Add 5/2 to both sides:
y = 3 or y-5/2 = -1/2

Substitute back for y = e^(2 x):
e^(2 x) = 3 or y-5/2 = -1/2

Take the natural logarithm of both sides:
2 x = log(3) or y-5/2 = -1/2

Divide both sides by 2:
x = (log(3))/2 or y-5/2 = -1/2

Add 5/2 to both sides:
x = (log(3))/2 or y = 2

Substitute back for y = e^(2 x):
x = (log(3))/2 or e^(2 x) = 2

Take the natural logarithm of both sides:
x = (log(3))/2 or 2 x = log(2)

Divide both sides by 2:
Answer: | 
| x = (log(3))/2 or x = (log(2))/2

Melodie: You made a mistake by making it=6 instead of -6

Also, when I substitute both of my answers, the equation balances, while your ln(6) does not.

Thanks.