Question

Least positive integral value of n for which $\frac{9^n}{{10}^n}<\frac{1}{2}$ is

Answer 1

We have $\frac{9^n}{{10}^n}<\frac{1}{2}$

$\Rightarrow 2\times 9^n<{10}^n$

Taking log on both sides

$\log 2\times 9^n<\log {10}^n\Rightarrow \log 2+n\log 9<n$

$\Rightarrow \log 2<n(1-2\log 3)$

$\therefore n>\frac{\log 2}{1-2\log 3}=\frac{0.3010}{1-2\left(0.4771\right)}=\frac{0.3010}{0.04575}=6.579$

Least integral value of $n$ is $7$