Question

If ${\log }_{a}3=2$ and ${\log }_{b}8=3$, then value of ${\log }_{a}b$ is equal to

• A. ${\log }_{3}2$
• B. ${\log }_{2}3$
• C. ${\log }_{3}4$
• D. ${\log }_{4}3$

Answer 1

The correct option is C ${\log }_{3}4$

We have, ${\log }_{a}3=2$ and ${\log }_{b}8=3$
By change of base rule, $({\log }_{a}b=\frac{\log b}{\log a})$,
$\frac{\log 3}{\log a}=2$ and $\frac{\log 8}{\log b}=3$
$\Rightarrow \log a=\frac{\log 3}{2}$ and $\log b=\frac{\log 8}{3}=\frac{3\log 2}{3}=\log 2$
Substituting these values in ${\log }_{a}b$, we get
$\therefore {\log }_{a}b=\frac{\log b}{\log a}=\frac{\log 2}{\frac{\log 3}{2}}=\frac{2\log 2}{\log 3}=\frac{\log 4}{\log 3}={\log }_{3}4$