Question

logarithm question

Answer 1

\(\large \color{black}{\begin{align} &\text{Prove}\ \\~\\& \log_{0.625} \sqrt{128}=\dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

http://www.wolframalpha.com/input/?i=is+%5Clog_%7B0.625%7D+%5Csqrt%7B128%7D%3D%5Cdfrac%7B2%2B%5Clog_%7B8%7D+2%7D%7B2%28%5Clog_%7B8%7D+5-1%29%7D+%3F

Answer 3

\[\begin{align}
\dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)}&=\dfrac{\log_8 8^2+\log_{8} 2}{2(\log_{8} 5-1)}\\~\\
&=\dfrac{\log_{8} 128}{2(\log_{8} 5-1)}\\~\\
&=\dfrac{\log_{8} \sqrt{128}}{\log_{8} 5-1}\\~\\
&=\cdots
\end{align}\]

Answer 4

how do i solve denominator

Answer 5

and base 0.625

Answer 6

inferior solution but what the hell:

\(0.625 = \frac{5}{8}\)

so you can base shift the LHS into \(log_8\)'s and do it the long way

Answer 7

that looks useful.

Answer 8

yes that would work change the base of the log so they are all the same

Answer 9

ok i got it now

Answer 10

thnx

Answer 11

\(\large log_{0.625}\sqrt128 = log_{\frac{5}{8}}\sqrt128 = \frac{log_8\sqrt{128}}{log_8\frac{5}{8}} = \frac{log_8\sqrt{128}}{log_8 5 - 1}\)

then some more fiddling about with \(\sqrt128 \)

sorry i actually latexed this so i am going to post it ..... even though i think you have killed this off :p

good luck!