mathmath333
  • mathmath333
logarithm question
Mathematics
chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} &\text{Prove}\ \\~\\& \log_{0.625} \sqrt{128}=\dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)} \hspace{.33em}\\~\\ \end{align}}\)
mathmath333
  • mathmath333
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
ganeshie8
  • ganeshie8
\[\begin{align} \dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)}&=\dfrac{\log_8 8^2+\log_{8} 2}{2(\log_{8} 5-1)}\\~\\ &=\dfrac{\log_{8} 128}{2(\log_{8} 5-1)}\\~\\ &=\dfrac{\log_{8} \sqrt{128}}{\log_{8} 5-1}\\~\\ &=\cdots \end{align}\]

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mathmath333
  • mathmath333
how do i solve denominator
mathmath333
  • mathmath333
and base 0.625
IrishBoy123
  • IrishBoy123
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
mathmath333
  • mathmath333
that looks useful.
cwrw238
  • cwrw238
yes that would work change the base of the log so they are all the same
mathmath333
  • mathmath333
ok i got it now
mathmath333
  • mathmath333
thnx
IrishBoy123
  • IrishBoy123
\(\large log_{0.625}\sqrt128 = log_{\frac{5}{8}}\sqrt128 = \frac{log_8\sqrt{128}}{log_8\frac{5}{8}} = \frac{log_8\sqrt{128}}{log_8 5 - 1}\) 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!

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