Solving for n. Logarithm

Question

How come:

$\dfrac{3^n}{2^{2n-1}}<0.5\cdot10^{-5}$

will be equal to $n = 45$?

Answer 1

$$\frac{3^n}{2^{2n-1}}=\frac{3^n}{\frac{2^{2n}}{2}}=2\cdot{\frac{3^n}{4^n}}=2\left(\frac{3}{4}\right)^n < \frac{1}{2}\cdot 10^{-5} \Rightarrow \left(\frac{3}{4}\right)^n < \frac{1}{4}\cdot{10^-5} \\ n\cdot{\log\left(\frac{3}{4}\right)}<\log\left(\frac{10^{-5}}{4}\right) \quad \text{Taking logarithm of both sides} \\ n > \frac{\log\left(\frac{10^{-5}}{4}\right)}{\log\left(\frac{3}{4}\right)}\approx 44.8 \quad \text{Dividing by }\log\left(\frac{3}{4}\right) \text{ which is negative} $$Hence, assuming $n\in\mathbb{N}$, the minimal value of $n$ that the inequality holds for is $45$.