Question

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

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

Answer 1

The correct option is D $\frac{1}{2ab-1}$

${\log }_{4}5=a\Rightarrow 4^a=5{\log }_{5}6=b\Rightarrow 5^b=6\Rightarrow 6=(4^a{)}^b=2^{2ab}\Rightarrow 3=\frac{2^{2ab}}{2}\Rightarrow 3=2^{2ab-1}$
$\text{Let}{\log }_{3}2=x\Rightarrow 3^x=2\cdots \left(1\right)\Rightarrow 3^x=2^{(2ab-1)x}\cdots \left(2\right)$
$\text{From}\left(1\right)\text{and}\left(2\right)(2ab-1)x=1x=\frac{1}{2ab-1}$