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mathmath333

  • one year ago

logarithm question

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} &\text{Prove}\ \\~\\& \log_{0.625} \sqrt{128}=\dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)} \hspace{.33em}\\~\\ \end{align}}\)

  2. ganeshie8
    • one year ago
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    \[\begin{align} \dfrac{2+\log_{8} 2}{2(\log_{8} 5-1)}&=\dfrac{\log_8 8^2+\log_{8} 2}{2(\log_{8} 5-1)}\\~\\ &=\dfrac{\log_{8} 128}{2(\log_{8} 5-1)}\\~\\ &=\dfrac{\log_{8} \sqrt{128}}{\log_{8} 5-1}\\~\\ &=\cdots \end{align}\]

  3. mathmath333
    • one year ago
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    how do i solve denominator

  4. mathmath333
    • one year ago
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    and base 0.625

  5. IrishBoy123
    • one year ago
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    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

  6. mathmath333
    • one year ago
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    that looks useful.

  7. cwrw238
    • one year ago
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    yes that would work change the base of the log so they are all the same

  8. mathmath333
    • one year ago
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    ok i got it now

  9. mathmath333
    • one year ago
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    thnx

  10. IrishBoy123
    • one year ago
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    \(\large log_{0.625}\sqrt128 = log_{\frac{5}{8}}\sqrt128 = \frac{log_8\sqrt{128}}{log_8\frac{5}{8}} = \frac{log_8\sqrt{128}}{log_8 5 - 1}\) 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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