Question

If ${\log }_{a}b=2,{\log }_{b}c=2$ and ${\log }_{3}c=3+{\log }_{3}a,$ then the value of $a+b+c$ is

• A. $90$
• B. $93$
• C. $102$
• D. $243$

Answer 1

The correct option is B $93$

Given: ${\log }_{a}b=2,{\log }_{b}c=2$ and ${\log }_{3}c=3+{\log }_{3}a$
For the $\log$ to be defined,
$a,b,c>0$
and $a\ne 1,b\ne 1$

${\log }_{a}b=2\Rightarrow b=a^2\dots \left(1\right){\log }_{b}c=2\Rightarrow c=b^2$
Using equation $\left(1\right),$ we get
$\Rightarrow c=a^4$

Now, ${\log }_{3}c=3+{\log }_{3}a$
$\Rightarrow {\log }_3{a}^4=3+{\log }_{3}a\Rightarrow 3{\log }_{3}a=3\Rightarrow a=3\Rightarrow b=9,c=81\therefore a+b+c=93$