Question

Prove that: $\frac{1}{3}<{\log }_{20}3<\frac{1}{2}$

Answer 1

Let $y={\log }_{20}3=\frac{1}{{\log }_{3}20}$ ($\because {\log }_{a}b=\frac{1}{{\log }_{b}a}$)

Since, $9<20<27$

$\Rightarrow {\log }_{3}9<{\log }_{3}20<{\log }_{3}27$

(When the base of logarithm is greater than $1$ and $m<n\Rightarrow {\log }_{a}m<{\log }_{a}n$)

$\Rightarrow {\log }_3{3}^2<{\log }_{3}20<{\log }_3{3}^3$

$\Rightarrow 2{\log }_{3}3<{\log }_{3}20<3{\log }_{3}3$ $(\log x^m=m\log x)$

$\Rightarrow 2<\frac{1}{y}<3$ ($\because {\log }_{a}a=1$)

$\Rightarrow \frac{1}{3}<y<\frac{1}{2}$

$\therefore \frac{1}{3}<{\log }_{20}3<\frac{1}{2}$