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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.


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.


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For real $x$ both $e^{0.5x} \gt 0 $ and $324e^{-0.5x} \gt 0$ – Henry Jun 4 '12 at 21:39
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) – Abdelmajid Khadari Jun 4 '12 at 21:43
up vote 2 down vote accepted

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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Well the question is coming from my homework here so I know it's not a typo. It's asking me solve for the x coordinates. So naturally, I'd set the equation to 0 and solve for x. – ninja08 Jun 4 '12 at 21:45
Then, if you're asking for real solutions, you can see from $z^2=-324$ that there are none. Complex ones are, of course. Be careful when taking the complex logarithm though, it is not uniquely defined. – Gregor Bruns Jun 4 '12 at 21:48
@ninja08: You might want to give the complete homework problem. There may be an error in how you got to the equation you posted. – André Nicolas Jun 4 '12 at 21:49
I didn't realize one of the answers was "There are no x-intercepts" lol Thanks! – ninja08 Jun 4 '12 at 22:12

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. – ninja08 Jun 4 '12 at 21:48

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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I see that now. Thanks for your help. – ninja08 Jun 4 '12 at 21:48
not at all, you are welcome. – Abdelmajid Khadari Jun 4 '12 at 21:51

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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Wait, what? $\ln(ab) = \ln a + \ln b$, not $(\ln a)(\ln b)$! – Rahul Jun 4 '12 at 21:45
@RahulNarain: I got into the spirit of the thing. Fixed. – André Nicolas Jun 4 '12 at 21:48

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