2
$\begingroup$

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

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 $\,-7\,$?

$\endgroup$
  • 2
    $\begingroup$ Not only is it possible: it actually is that. $\endgroup$ – DonAntonio Mar 29 '13 at 14:17
2
$\begingroup$

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

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.

$\endgroup$
  • $\begingroup$ 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 :-) $\endgroup$ – Jean-Claude Arbaut Mar 29 '13 at 14:29
  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Mar 30 '13 at 1:38
1
$\begingroup$

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 $10^y = \sqrt{10}$...

$\endgroup$
0
$\begingroup$

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

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 $\,\log=\log_{10}\,$

$\endgroup$
0
$\begingroup$

Formulas regarding log:

$$log_a(M\times N)=log_aM+\ log_aN\\log_a(M^n)=nlog_aM\\log_aM=\dfrac{log_tM}{log_ta};\forall a,t\not=1$$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.