Smallest integer in inequality

We want $4^n > 17^{19}$. Taking logarithms on both sides: $n \log 4 > 19 \log 17$ $n > 19 \frac{\log 17}{\log 4} = 19 \log_4 17$

Note that $17 \approx 16 = 4^2$, so $\log_4 17 \approx 2$. More precisely, $17 = 16 \times 1.0625 = 2^4 \times 1.0625$. Using $\log_{10} 2 \approx 0.3010, \log_{10} 17 \approx 1.2304$: $\log_4 17 = \frac{1.2304}{2 \times 0.3010} = \frac{1.2304}{0.6020} \approx 2.0438$

$n > 19 \times 2.0438 \approx 38.83$ Thus, the smallest integer $n = 39$, which is closest to 39.