The value of N = 2 log_{_{6}}^{3} 3 \textrm{ } \left(\right. 1 + 3 log_{_{3}}^{2} 2 \left.\right) - log_{_{6}}^{3} \left(\right. \frac{3}{2} \left.\right) is equal to

Engineering

Mathematics

Properties of Logarithm

Question

The value of $N = 2 \log_{_{6}}^{3} 3 (1 + 3 \log_{_{3}}^{2} 2) - \log_{_{6}}^{3} (\frac{3}{2})$ is equal to

–1

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$\frac{2}{{(\log_{_{3}} 6)}^{3}} (1 + 3 \log_{_{3}}^{2} 2) - \frac{{(1 - \log_{_{3}} 2)}^{3}}{{(1 + \log_{_{3}} 2)}^{3}}$

Let log₃2 = t

$= \frac{2 (1 + 3 t^{2})}{{(1 + t)}^{3}} - \frac{{(1 - t)}^{3}}{{(1 + t)}^{3}}$

$= \frac{2 + 6 t^{2} - 1 + t^{3} + 3 t - 3 t^{2}}{{(1 + t)}^{3}}$

$= \frac{t^{3} + 3 t + 3 t^{2} + 1}{{(1 + t)}^{3}} = 1$