If {log ₄}5 = a and {log ₅}6 = b, then {log ₃}2 is equal to

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Basic of Logarithms

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If ${\log _4}5 = a$ and ${\log _5}6 = b,$ then ${\log _3}2$ is equal to

A

${1 \over {2a + 1}}$

B

${1 \over {2b + 1}}$

C

$2ab + 1$

D

${1 \over {2ab - 1}}$

Solution

(d) $ab = {\log _4}5.{\log _5}6 = {\log _4}6 = {1 \over 2}{\log _2}6$

$ab = {1 \over 2}(1 + {\log _2}3) \Rightarrow 2ab – 1 = {\log _2}3$

$\therefore {\log _3}2 = {1 \over {2ab – 1}}$.

Standard 11

Mathematics

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