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anonymous

  • one year ago

Help please: log6(1/5√36)

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  1. anonymous
    • one year ago
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    \[\log_{6} \frac{ 1 }{ \sqrt[5]{36} }\]

  2. anonymous
    • one year ago
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    So far I've done: |dw:1442966236088:dw|

  3. jdoe0001
    • one year ago
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    so... what are you asked to do with \(\bf log_{6}\left( \cfrac{ 1 }{ \sqrt[5]{36} }\right) ?\)

  4. anonymous
    • one year ago
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    Find x.

  5. jdoe0001
    • one year ago
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    "x"? kinda hard to find, since it isn't in the expression firstly

  6. jdoe0001
    • one year ago
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    ohh hmm just saw it ... so is \(\bf \bf log_{6}\left( \cfrac{ 1 }{ \sqrt[5]{36} }\right) =x\) then... ok

  7. anonymous
    • one year ago
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    Yup, says evaluate it. My notes said to set each equation to x and solve for it.

  8. jdoe0001
    • one year ago
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    well.... can we say...turn \(\large \bf \sqrt[5]{36}\) into a a value with a rational exponent? notice \(\large \sqrt[5]{36}\implies \sqrt[5]{6^2}\)

  9. jdoe0001
    • one year ago
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    recall that \(\large { a^{\frac{{\color{blue} n}}{{\color{red} m}}} \implies \sqrt[{\color{red} m}]{a^{\color{blue} n}} \qquad \qquad \sqrt[{\color{red} m}]{a^{\color{blue} n}}\implies a^{\frac{{\color{blue} n}}{{\color{red} m}}} \\\quad \\ % rational negative exponent a^{-\frac{{\color{blue} n}}{{\color{red} m}}} = \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}} \implies \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}\qquad\qquad % radical denominator \cfrac{1}{\sqrt[{\color{red} m}]{a^{\color{blue} n}}}= \cfrac{1}{a^{\frac{{\color{blue} n}}{{\color{red} m}}}}\implies a^{-\frac{{\color{blue} n}}{{\color{red} m}}} }\)

  10. anonymous
    • one year ago
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    That makes sense. So now: 6^x = 6^2/5. Sixes cross out right? x = 2/5? Negative, since it was a fraction.

  11. jdoe0001
    • one year ago
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    close... one sec

  12. jdoe0001
    • one year ago
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    \(\bf log_{6}\left( \cfrac{ 1 }{ \sqrt[5]{36} }\right) =x\implies log_6\left( \cfrac{1}{6^{\frac{2}{5}}} \right)=x\implies log_{\color{red}{ 6}}\left( {\color{red}{ 6}}^{-\frac{2}{5}} \right)=x \\ \quad \\ \textit{log cancellation rule } \qquad log_{\color{red}{ a}}{\color{red}{ a}}^x\implies x\qquad \qquad {\color{red}{ a}}^{log_{\color{red}{ a}}x}=x\qquad thus \\ \quad \\ -\cfrac{2}{5}=x\)

  13. anonymous
    • one year ago
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    Thanks!

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