# Solve e^(2 x)-e^x-12 = 0 for x

1. Dec 9, 2011

### physicsdreams

1. The problem statement, all variables and given/known data
solve e^(2 x)-e^x-12 = 0 for x

2. Relevant equations

3. The attempt at a solution

We had this question on a test, and after solving it (i'd rather not write out the work), I came to two solutions.
Ln(4)
Ln(-3)

I plugged both answers back into the equation, and they both worked.
My teacher says that Ln(-3) is NOT an answer, because it is an "extraneous solution".
I did some messing around on Wolframalpha, and they seem to agree that ln(-3) is, in fact, a solution.

Can someone please clarify if both answers are in fact solutions, or if one is an extraneous solution. Or if you think he meant something else, please tell me. please keep in mind that i'm still in precalculus, so please explain it in as simple terms as possible.

Thanks.

2. Dec 9, 2011

### micromass

You have defined the logarithm only for positive values of x. So log(-1) is not defined. The logarithm is a function that takes in positive values and spews out a real number.

It is possible to define the logarithm for negative numbers, but that's not what your teacher wants. Your teacher works with the real logarithm, and thus is not defined for negative numbers.

3. Dec 10, 2011

### physicsdreams

I've now talked to my teacher, and he thinks that I might be right; or at least he isn't sure why plugging in the solutions back into the equations work.

If we are talking about logarithms that are defined by both positive and negative numbers (let's just say the problem allowed for all possible answers), are both proposed answer(s) 'right'.

Is the answer considered 'right' if you can plug it in?

Thanks

Last edited: Dec 10, 2011
4. Dec 10, 2011

### BloodyFrozen

Again, $ln(-3)$ is a root if logarithms are defined for negative numbers.
In your case, I'm sure you're working with $\{x\in \mathbb{R} :x > 0\}$
$$ln(-1)=i \pi$$
$$ln(-3)=ln(3)+i \pi$$

Complex numbers are introduced when you take the logarithm of a negative number.

Last edited: Dec 10, 2011