Simplify a logarithm function

$$2\log\sqrt[4]{10}-\ln e^{-7}+\log_9\sqrt 3$$

I want to simplify this function. I believe that $\,2\log\sqrt[4]{10}\,$ can become $\,\log\sqrt{10}\,$ but now I'm stuck.

Is it possible that $\ln e^{-7}\,$ can be just $\,-7\,$?

• Not only is it possible: it actually is that. – DonAntonio Mar 29 '13 at 14:17

$$\log\sqrt[4]{10}= \frac14\log10$$ $$\ln e^{-7}=-7\ln e = -7\cdot1$$ $$\log_9\sqrt{3} = \frac12\log_9 3=\frac12\log_9\sqrt{9}=\frac12\cdot\frac12=\frac14$$

Probably "$\log10$" is intended to mean $\log_{10}10$, so that is equal to $1$.

I disapprove of using "$\log$" with no base to mean $\log_{10}$ when there's no special context saying that's the right base to use.

• The main reason to disapprove is that nowadays one usually means $\log = \ln$, whereas until some years ago, when slide rules and tables of logarithms were still widespread, usually one meant $\log=\log_{10}$. By the way, as I still use these tools myself (for fun, I confess, a kind of masochistic fun), I prefer the $\log=\log_{10}$ convention :-) – Jean-Claude Arbaut Mar 29 '13 at 14:29
• In some fields, maybe biology, chemistry, astronomy, etc., base-10 logarithms are still used, and they write "log" with no base for those. In some other fields (mathematics, statistics, computer science) "log" with no base means base-$e$. It's context-dependent, and at least that fact should be mentioned. – Michael Hardy Mar 30 '13 at 1:38

You are on the right path. Also, note that if $\log_9 (\sqrt{3}) = x$, then $9^x = \sqrt{3}$.

And if $\log = \log_{10}$, then you may be able to simplify $\log_{10} \sqrt{10}$ as well. In that case, you'd be looking for a number $y$ satisfying $10^y = \sqrt{10}$...

$$2\log\sqrt[4]{10}-\ln e^{-7}+\log_9\sqrt 3=\frac{2}{4}\log{10}-(-7)+\frac{1}{2}\log_93=\frac{1}{2}+7+\frac{1}{4}\ldots$$

Properties used:

$$\log_aa^n=n\;\;,\;\;\log_ax^n=n\log_ax$$

and, of course, the very definition of logarithm and the assumed fact that you surely meant $\,\log=\log_{10}\,$

Formulas regarding log:

Basically $log_aM=N$ means that $a^N=M$

Now the problem is easy,Ihope you'll be able to solve it yourself.