Question

If $lo{g}_{3}X=lo{g}_{9}16,$ then X+1=

Answer 1

Given:- ${\log }_{3}X={\log }_{9}16$

To find:- $X+1=?$

${\log }_{3}X={\log }_{9}16$

$\Rightarrow {\log }_{3}X={\log }_{3^2}16$

$\Rightarrow {\log }_{3}X=\frac{1}{2}{\log }_{3}16(\because {\log }_{a^n}b=\frac{1}{n}{\log }_{a}b)$

$\Rightarrow 2{\log }_{3}X={\log }_{3}16$

$\Rightarrow {\log }_3{X}^2={\log }_{3}16(\because \log a^n=n\log a)$

$\Rightarrow X^2=16(\because {\log }_{a}b={\log }_{a}c\Rightarrow b=c)$

$\Rightarrow X=\sqrt{16}=4$

Now,

$X+1=4+1=5$

Thus the value of $X+1$ is $5$.

Hence the correct answer is $5$.