Question

If $N={\mathrm{antilog}}_7\left({\log }_3\left({\mathrm{antilog}}_{\sqrt{3}}\left({\log }_{3}81\right)\right)\right)$, then ${\log }_{3}N$ lies between two successive integers $a$ and $b$. Find $(a+b)$.

Answer 1

Answer: 7

$N={\mathrm{antilog}}_7\left({\log }_3\left({\mathrm{antilog}}_{\sqrt{3}}\left({\log }_{3}81\right)\right)\right)={\mathrm{antilog}}_7\left({\log }_3(\sqrt{3}{)}^4\right)$
$={\mathrm{antilog}}_7\left({\log }_{3}9\right)={\mathrm{antilog}}_7(2)=7^2=49$
$N=49\Rightarrow 3<{\log }_{3}43<4\Rightarrow a+b=7$