Question

If ${\log }_{a}b=2;{\log }_{b}c=2$ and ${\log }_{3}c=3+{\log }_{3}a$ then $(a+b+c)$ equals

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

Answer 1

The correct option is B $93$

$lo{g}_{a}b=2$

$⟹b=a^2$

$lo{g}_{b}c=2$

$⟹c=b^2$

$lo{g}_{3}c=3+lo{g}_{3}a$

$⟹lo{g}_{3}c-lo{g}_{3}a=3$

Using identity

$lo{g}_{p}x-lo{g}_{p}y=lo{g}_p\frac{x}{y}$ we get

$lo{g}_3\left(\frac{c}{a}\right)=3$

$\frac{c}{a}=3^3$

$\frac{c}{a}=27$

$c=27a$

$b^2=27a$

$(a^2{)}^2=27a$

$a^4-27a=0$

$a(a^3-27)=0$

$a^3-27=0$

$a=3$

Putting this value we will get

$b=3^2$

$b=9$

Putting this value we will get

$c=9^2$

$c=81$

$a+b+c=3+9+81=93$