183 views

1 vote

Given that, $\log_{3}{5}=\log_{5}{(x+2)} ;x\in \mathbb{R}$

We know that, $\log_{3}{5}>\log_{3}{3}$

$\Rightarrow \boxed{\log_{3}{5}>1} \quad [\because \log_{a}a = 1]$

And, $\log_{3}{5}<\log_{3}{9}$

$\Rightarrow \log_{3}{5}<\log_{3}{3^{2}}$

$\Rightarrow \log_{3}{5} < 2 \log_{3}{3} \quad [\because \log_{b}a^{x} = x \log_{b}a]$

$\Rightarrow \boxed{\log_{3}{5}<2}$

So, $ \boxed{1<\log_{3}{5}<2}$

Left-hand side lies between $1$ and $2$. So, the right-hand side also follows the same.

Now, $ 1<\log_{5}{(x+2)}<2$

$\Rightarrow 5^{1}<x+2<5^{2} \quad \left[\because \log_{a}{x}=b \Rightarrow x=a^{b}\right]$

$\Rightarrow 5<x+2<25 $

$\Rightarrow \boxed{3<x<23}$

Correct Answer $:\text{B}$

We know that, $\log_{3}{5}>\log_{3}{3}$

$\Rightarrow \boxed{\log_{3}{5}>1} \quad [\because \log_{a}a = 1]$

And, $\log_{3}{5}<\log_{3}{9}$

$\Rightarrow \log_{3}{5}<\log_{3}{3^{2}}$

$\Rightarrow \log_{3}{5} < 2 \log_{3}{3} \quad [\because \log_{b}a^{x} = x \log_{b}a]$

$\Rightarrow \boxed{\log_{3}{5}<2}$

So, $ \boxed{1<\log_{3}{5}<2}$

Left-hand side lies between $1$ and $2$. So, the right-hand side also follows the same.

Now, $ 1<\log_{5}{(x+2)}<2$

$\Rightarrow 5^{1}<x+2<5^{2} \quad \left[\because \log_{a}{x}=b \Rightarrow x=a^{b}\right]$

$\Rightarrow 5<x+2<25 $

$\Rightarrow \boxed{3<x<23}$

Correct Answer $:\text{B}$