Question

Let ${\log }_{3}2=a$ then find the value of ${\log }_{8}27+{\log }_{27}8$ in terms of $a$.

Answer 1

Given, ${\log }_{3}2=a\dots \dots \left(i\right)$.
Now, ${\log }_{8}27+{\log }_{27}8$
$={\log }_{2^3}{3}^3+{\log }_{3^3}{2}^3$
$=\frac{3}{3}{\log }_{2}3+\frac{3}{3}{\log }_{3}2(\because lo{g}_{a^m}{b}^n=\frac{n}{m}lo{g}_{a}b)$
$=\frac{1}{{\log }_{3}2}+{\log }_{3}2(\because lo{g}_{n}m=\frac{1}{lo{g}_{m}n})$
$=\frac{1}{a}+a$. [ Using (i)]