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The smallest integer $n$  for which $4^{n}>17^{19}$ holds, is closest to

1. $33$
2. $37$
3. $39$
4. $35$

## 1 Answer

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 $:\text{C}$

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