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 $\,-7\,$?
 A: $$\log\sqrt[4]{10}= \frac14\log10$$
$$\ln e^{-7}=-7\ln e = -7\cdot1$$
$$\log_9\sqrt{3} = \frac12\log_9 3=\frac12\log_9\sqrt{9}=\frac12\cdot\frac12=\frac14$$
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.
A: 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 $10^y = \sqrt{10}$...
A: $$2\log\sqrt[4]{10}-\ln e^{-7}+\log_9\sqrt 3=\frac{2}{4}\log{10}-(-7)+\frac{1}{2}\log_93=\frac{1}{2}+7+\frac{1}{4}\ldots$$
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 $\,\log=\log_{10}\,$
A: 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.
