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I solved this problem in my textbook but noticed their solution was different than mine.

$1. \ 9e^{-2x}=1$

$2. \ e^{-2x}=\frac{1}{9}$

$3. -2x=\ln(\frac{1}{9})$

$4. \ x=-\ \frac{1}{2}\ln(\frac{1}{9})$

However, the answer that my textbook gives is $\frac{\ln(9)}{2}$

I plugged these expressions into my calculator and they are indeed equivalent, however I don't see what properties I could use to get from my messy answer to the textbook's much cleaner one. Any help would be greatly appreciated. Thank you.

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    Thanks to all for the immediate multitude of answers! Truly shows how great this community is. I will accept S.C.B's as it also provides an additional proof of the property that I lacked knowledge of. – 2017-02-16
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    ln(1/9) is - ln(9) – 2017-02-16

4 Answers 4

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Note that $$\ln x +\ln y =\ln xy, \; \ln 1=\ln e^{0}=0$$ If $x, y$ are positive reals, as seen here. From this, $$\ln x +\ln \frac{1}{x}=0 \iff \ln x =-\ln \frac{1}{x}$$ So $$\ln \frac{1}{9}=-\ln 9$$ So $-\frac{1}{2}\ln(\frac{1}{9})=\frac{\ln(9)}{2}$

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    Thank you for not only clearing up which property I lacked but also demonstrating why it works! This clears up any confusion I had. – 2017-02-16
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    @HarnoorLal You're welcome. Glad to help. In general, it helps to know $\ln x=-\ln x^{-1}$. But to be shorter, it follows from the defenition of exponentiation that $$\ln x^{n}=n \ln x$$ – 2017-02-16
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    +1 For showing a clear and adequate proof of why this works. – 2017-02-16
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    An intuitive notion of log is x to the power of what becomes y? Say the number is z and y is changed to 1/y. Then obviously z becomes -z. – 2017-02-16
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    @JimThio Isn't that the definition of $\log$, at least in high school? (I don't know enough math to know if it always the definition). – 2017-02-16
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    @HarnoorLal: Be careful. This answer is imprecise and the first claim does not hold if you put $x = y = -1$. – 2017-02-16
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    @user21820 I edited that in. – 2017-02-16
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    Great! Thanks for adding the correct conditions! =) – 2017-02-16
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    @user21820 Yeah, I really shout not have forgotten that. Thank for reminding me and helping me make a more accurate answer. – 2017-02-16
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    @HarnoorLal I edited the answer to accomodate the error user21820 kindly pointed out to me. – 2017-02-16
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$\ln (1/9) = \ln (9^{-1})=-1 \cdot \ln (9)$

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There exists the following property for logarithms:

$$n \ln{x} = \ln{x^n}$$

So for your problem you have:

$$ -\frac{1}{2} \ln{\left(\frac{1}{9}\right)}=\frac{1}{2}\ln{\left(\left(\frac{1}{9}\right)^{-1}\right)}=\frac{1}{2}\ln{9}= \frac{\ln9}{2}$$

I hope this is sufficient as an explanation.

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    Please state the precise conditions for your claim. Namely, what is $n$ and $x$? It does not hold for all pairs of real numbers. – 2017-02-16
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$\ln(\frac{1}{9})=\ln(9^{-1})=(-1)\ln(9)$ ;)