Question

If ${log}_{4}5=aand{log}_{5}6=b,then{log}_{3}2$ is equal to

• A. $\frac{1}{2a+1}$
• B. $\frac{1}{2b+1}$
• C. $2ab+1$
• D. $\frac{1}{2ab-1}$

Answer 1

Answer: (D)

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

Given, ${\log }_{4}5=a$ and ${\log }_{5}6=b$

Now, $ab={\log }_{4}5\times {\log }_{5}6$

$⟹ab=\frac{\log 5}{\log 4}\times \frac{\log 6}{\log 5}$

$⟹ab=\frac{\log 6}{\log 4}$

$⟹ab={\log }_{4}6$

$⟹ab=\frac{1}{2}{\log }_{2}6$

$⟹ab=\frac{1}{2}{\log }_2(2\times 3)$

$⟹ab=\frac{1}{2}({\log }_{2}2+{\log }_{2}3)$

$⟹ab=\frac{1}{2}(1+{\log }_{2}3)$

$⟹2ab=1+{\log }_{2}3$

$⟹{\log }_{2}3=2ab-1$

$\therefore {\log }_{3}2=\frac{1}{2ab-1}$.