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Algebra Level 3

$\left ( \log_{\sqrt{6}} 3 \right )^2 + \frac {1}{ \log_2 \sqrt{6} \cdot \log_{18} \sqrt{6} } = \ ?$

1 solution

Noel Lo
Mar 30, 2015

$(log_{\sqrt{6}}{3})^2$ + $\frac{1}{log_{2} \sqrt{6} * log_{18} \sqrt{6}}$

= $(log_{\sqrt{6}}{3})^2$ + $\log_{\sqrt6}{2}$ * $\log_{\sqrt{6}}{18}$

= $(log_{\sqrt{6}}{3})^2$ + $\log_{\sqrt6}(\frac{6}{3})$ * $\log_{\sqrt{6}}(6 \times3)$

= $(log_{\sqrt{6}}{3})^2$ + ( $\log_{\sqrt6}{6} - log_{\sqrt6}{3}$ ) ( $\log_{\sqrt6}{6} + log_{\sqrt6}{3}$ )

= $(log_{\sqrt{6}}{3})^2$ + $(2- log_{\sqrt6}{3}$ ) ( $2 + log_{\sqrt6}{3}$ )

= $(log_{\sqrt{6}}{3})^2$ + $2^2$ - $(log_{\sqrt{6}}{3})^2$ = $\boxed{4}$