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prove that :$\frac{1}{\log _{3} 6}+\frac{1}{\log _{8} 6}+\frac{1}{\log _{9} 6}=3$

written 5.6 years ago by teamques10 ★ 69k

modified 3.0 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.6 years ago by teamques10 ★ 69k

Solution:

$$\begin{aligned} \text {L.H.S} &=\frac{1}{\log _{3} 6}+\frac{1}{\log _{8} 6}+\frac{1}{\log _{9} 6} \ &=\frac{\log 3}{\log 6}+\frac{\log 8}{\log 6}+\frac{\log 9}{\log 6} \ &=\frac{\log (3 \times 8 \times 9)}{\log 6}\ & =\frac{\log 216}{\log 6}\ & =\frac{\log 6^{3}}{\log 6}\ & =\frac{3 \log 6}{\log 6} \ & =3=R \cdot H . S \end{aligned}$$

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