Basic of Logarithms
medium

જો ${\log _4}5 = a$ અને ${\log _5}6 = b $ તો ${\log _3}2= . . . .$

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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.