How to solve this equation involving log?

Question

I want to know for which natural numbers $n$ we have the inequality $n < 8\log_2(n)$. I know the answer is $n \leq 43,$ but I have no idea how to get there.

Answer 1

I'm just going to explain the answer of Dr. Sonnhard Graubner more thoroughly.

It is well known that $\log_2 n = \frac{\ln (n)}{\ln (2)}$

So we have $$n < \frac{8}{\ln (2)} \cdot \ln(n)$$ Which is equivalent to

$$\frac{n}{\ln(n)} < \frac{8}{\ln(2)}$$

Now consider the function $h(x)=\frac{x}{\ln(x)}$, we have $h'(x)= \frac{\ln(x)-1}{\ln^2(x)}$

Therefore $h$ is increasing whenever $\ln(x)>1$, so for $x>e$. It follows that $\frac{n}{\ln(n)}$ is increasing for $n\geq 3$.

Therefore it it enough to find the first time $\frac{n}{\ln(n)}$ passes $\frac{8}{\ln(2)}$. Checking by hand (say using a calculator) you can show that for $n=43$ we have an inequality while for $n=44$ we don't.

Answer 2

You can write $$\frac{\ln(2)}{8}=\frac{\ln(n)}{n}$$ and consider the function $$h(n)=\frac{\ln(n)}{n}$$