Question

If ${\log }_{4}5=a$ and ${\log }_{5}6=b$, then the value of ${\log }_{3}2$ is

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

Answer 1

Answer: (D)

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

We have,
${\log }_{4}5=a$ and ${\log }_{5}6=b$
Now using, ${\log }_{y}x=\frac{\log x}{\log y}$, we get

$a=\frac{\log 5}{\log 4}=\frac{\log 5}{\log 2^2}=\frac{\log 5}{2\log 2},(\because \log a^x=x\log a)$
$\therefore \log 2=\frac{\log 5}{2a}$$\Rightarrow$(1)

$b=\frac{\log 6}{\log 5}=\frac{\log 2+\log 3}{\log 5},(\because \log (xy)=\log x+\log y)$

$\therefore \log 3=b\log 5-\log 2$
$\therefore \log 3=b\log 5-\frac{\log 5}{2a}=\frac{\log 5(2ab-1)}{2a}$ $\Rightarrow$(2)

Since $\log 2=\frac{\log 5}{2a}$

$\therefore {\log }_{3}2=\frac{\log 2}{\log 3}=\frac{\frac{\log 5}{2a}}{\frac{\log 5(2ab-1)}{2a}}=\frac{1}{2ab-1}$