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I had problems understanding how to solve $$ 6^{-\log_{6}^2} $$

Any help would be much appreciated.

Thanks!

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    $\begingroup$ can you tell me if the expression that you wrote is $\displaystyle 6^{-\log_6 2}$? $\endgroup$ – DiegoMath Jul 15 '14 at 22:29
  • $\begingroup$ I am guessing that it is. $\endgroup$ – Juanito Jul 15 '14 at 22:41
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Let, $$ x=6^{-\log_6 2} $$ Thus, $$ \log_6 x =-\log_6 2 = \log_6 2^{-1} = \log_6 \frac{1}{2} \quad \Rightarrow \quad x=\frac{1}{2} $$

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\begin{equation}e^x = y \rightarrow ln(y) = x \\ \rightarrow e^{ln(y)}= e^x=y \\ thus: \\ 6^{-log_6(2)}=({6^{log_6{2}}})^{-1}=(2)^{-1}=1/2 \end{equation}

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$$ \begin{align} 6^{-\log_6(2)} &= \frac{1}{6^{\log_6(2)}} \quad \text{by definition of negative power}\\ &= \frac{1}{2} \quad \text{since $6^x$ and $\log_6(x)$ are inverses} \end{align} $$

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