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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!

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    this equation is equivalent to $e^{x}+354e=0$ so $e^{x}=-354e$ and then you apply the $\ln$ to both sides, but her we have -354e<0 (just to know the methode of solving)2012-06-04

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this equation have no solution simply because $e^{0.5x}>0$ and $e^{-0.5x}>0$ for all $x$ in $\mathbb{R}$.

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    not at all, you are welcome.2012-06-04
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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 $x=\dfrac{\ln u}{0.5}$.

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    Yes I see how this works now. Thanks for your help.2012-06-04
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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?

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    @RahulNarain: I got into the spirit of the thing. Fixed.2012-06-04
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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.

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    I didn't realize one of the answers was "There are no x-intercepts" lol Thanks!2012-06-04