# e^(2x)+4e^x-12=0 find the value of x that satisfies the equation

• Dec 9th 2008, 10:11 AM
AlexTweed
e^(2x)+4e^x-12=0 find the value of x that satisfies the equation
By drawing a graph I find x is about 0.693... but I need to show how to work it out using an algebraic method. I think I'm stuck on trying to get one x on one side of the equation if that makes sense. Please help! thanks

• Dec 9th 2008, 10:22 AM
Chop Suey
As you can see, it is a quadratic. To make it clearer, let $u = e^x$, therefore:

$u^2+4u-12=0$

Now can you solve it?
• Dec 9th 2008, 10:25 AM
Soroban
Hello, Alex!

Quote:

Solve: . $e^{2x} + 4e^x - 12 \:=\:0$

Factor: . $\left(e^x+6\right)\left(e^x - 2\right) \;=\;0$

Then we have: . $\begin{Bmatrix}e^x+6 \:=\:0 & \Rightarrow & e^x \:=\:-6 & & \text{Impossible} \\ \\[-3mm]
e^x - 2 \:=\:0 & \Rightarrow & e^x \:=\:2 & \Rightarrow & \boxed{x \:=\:\ln(2)}\;\; \end{Bmatrix}$