# Thread: [SOLVED] Log equation problem.

1. ## [SOLVED] Log equation problem.

Hi guys,

I'm doing exercises with exponential and logarithmic equations and I got to this one:

e^x - 12e^-x - 1 = 0

I was tempted to factor a e^x from the first two terms but guess I can't do that? I don't recognize this as a quadratic and frankly I'm not sure how to go about it. Any help is appreciated!

2. Originally Posted by DannyMath
Hi guys,

I'm doing exercises with exponential and logarithmic equations and I got to this one:

e^x - 12e^-x - 1 = 0

I was tempted to factor a e^x from the first two terms but guess I can't do that? I don't recognize this as a quadratic and frankly I'm not sure how to go about it. Any help is appreciated!
You are almost right. You should factorise $\displaystyle e^{-x}$ instead.
$\displaystyle e^{-x} (e^{2x} - e^x - 12) = 0$

Then substitute $\displaystyle a = e^x$ to get a quadratic

3. Haha I've never factorised a negative exponent in such a manner! Ingenious! Thanks.

4. Another way to do that is to multiply both sides of $\displaystyle e^x- 12e^{-x}- 1= 0$ by $\displaystyle e^x$ to get $\displaystyle (e^x)^2- 12- e^x= 0$. Let $\displaystyle a= e^x$ and the equation becomes the quadratic $\displaystyle a^2- a- 12= (a- 4)(a+ 3)$.

Now you have $\displaystyle a= e^x= 4$ so x= ln(4) and $\displaystyle a= e^x= -3$ which is not true for any x.

5. Oh yes, that seems like a straightforward solution, thanks!