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mathmath333
 one year ago
logarithm question
mathmath333
 one year ago
logarithm question

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} &\text{Prove}\ \\~\\& \log_{0.625} \sqrt{128}=\dfrac{2+\log_{8} 2}{2(\log_{8} 51)} \hspace{.33em}\\~\\ \end{align}}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\begin{align} \dfrac{2+\log_{8} 2}{2(\log_{8} 51)}&=\dfrac{\log_8 8^2+\log_{8} 2}{2(\log_{8} 51)}\\~\\ &=\dfrac{\log_{8} 128}{2(\log_{8} 51)}\\~\\ &=\dfrac{\log_{8} \sqrt{128}}{\log_{8} 51}\\~\\ &=\cdots \end{align}\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1how do i solve denominator

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2inferior 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
 one year ago
Best ResponseYou've already chosen the best response.1that looks useful.

cwrw238
 one year ago
Best ResponseYou've already chosen the best response.2yes that would work change the base of the log so they are all the same

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(\large log_{0.625}\sqrt128 = log_{\frac{5}{8}}\sqrt128 = \frac{log_8\sqrt{128}}{log_8\frac{5}{8}} = \frac{log_8\sqrt{128}}{log_8 5  1}\) 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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