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Given that,

$ 4^{n} > 17^{19} $

Taking the $ \log_{10}$ both sides.

$ \log_{10}4^{n} > \log_{10} 17^{19} $

$ \Rightarrow n \log_{10}4 > 19 \log_{10}17 \quad [ \because \log_{b}a^{x} = x \log_{b}a] $

$ \Rightarrow n > 19 \left( \frac{\log_{10}17}{ \log_{10}4} \right) $

$ \Rightarrow n > 19 ( \log_{4}17) \quad \left[ \because \log_{b}a = \frac{ \log_{x}a}{ \log_{x}b} \right] $

For easy calculation, let us assume $17 \approx 16.$

Now, $ n > 19 ( \log_{4} 16)$

$ \Rightarrow n > 19 ( \log_{4} 4^{2})$

$ \Rightarrow n > 19 \times 2 ( \log_{4}4)$

$ \Rightarrow \boxed{n >38} \quad [ \because \log_{a}a = 1 ] $

$ \Rightarrow \boxed{n \simeq 39} $

$\therefore$ The small integer $n$ is closed to $39.$

$\textbf{Short Method :}$

Given that, $ 4^{n} > 17^{19} $

$ \Rightarrow \left( 4^{2} \right)^{\frac{n}{2}} > 17^{19} $

$ \Rightarrow (16)^{\frac{n}{2}} > 17^{19} $

Here, $ 16 < 17 ,$ so $ \frac{n}{2}$ must be greater than $19.$

Thus, $ \frac{n}{2} > 19 $

$ \Rightarrow n > 38 $

$ \Rightarrow \boxed{n \simeq 39} $

Correct Answer $:\text{C}$