e's with x in the exponents: Gotta solve for x. I usually get lost when there are these exponential questions. I'm not used to seeing them.
I must solve for x.
$$f(x)=e^{0.5x}+324e^{-0.5x}=0$$
If I do  $f(x)=\ln e^{0.5x}+\ln 324e^{-0.5x}=\ln 0$
$\ln0$ is undefined. 
If I move a term over the equal sign it becomes negative and $\ln \#$ is undefined as well. 
Thanks!
 A: In the first place, $\ln(A+B)$ is not $(\ln A) + (\ln B)$, so your first step is wrong.
Next $e$ raised to a real power is always positive, and you can't have two positive numbers adding up to $0$, so this has no solutions.
Finally, if you had a positive number rather than $0$ there, you could do this:
Let $u=e^{0.5x}$, so that $e^{-0.5x} = \dfrac 1 u$.
Then you've got
$$
u + 324\cdot\frac 1 u = \text{whatever},
$$
$$
u^2 + 324 = (\text{whatever} \cdot u).
$$
That's a quadratic equation, and you can solve it for $u$.
Once you know what $u$ is, then $\ln u = 0.5x$, so $x=\dfrac{\ln u}{0.5}$.
A: Your logarithm manipulation is not correct. The logarithm of a sum is not equal to the sum of the logarithms.  If $a$ and $b$ are positive, we have $\ln(ab)=\ln a+\ln b$, but $\ln(a+b)\ne \ln a+\ln b$.  
Anyway, for real $x$ the left side is positive, so your equation has no real root. Is there a typo?
A: The general idea in solving those kind of problems is to do a substitution $z=e^{ax}$, solve the resulting equation, then take the logarithm of the solution.
In your case, you could, for instance, set $z=e^{\frac{1}{2}x}$. Your equation becomes then
$$z+324\frac{1}{z} = 0$$
Solving that for $z$ gives
$$z^2 = -324$$
and from that you get $z=\pm 18i$.
To get $x$ you would have to take logarithms:
$$\ln(\pm 18i) = \frac{1}{2}x$$
I suppose, though, that you had a typo in your question, some sign error maybe. You would get real solutions for $-324$ instead of $324$, for instance.
A: this equation have no solution simply because $e^{0.5x}>0$ and $e^{-0.5x}>0$ for all $x$ in $\mathbb{R}$.
