Is there an intersection point for $f_1 = 100n²$ and $f_2 = 2ⁿ$?

Question

As far as i know, there is, and it can be answered by resolving the equation $100n^2 = 2^n$.

But how can it be resolved? I already tried to transform the equation using $\log 2$, but i just don't get the point.

Can you please show me, how it's done?

Answer 1

Consider $f(n)=2^n-100n^2$. Note that this function is continuous, $f(1)<0$, and $f(100)>0$. Thus there is some point $c$ between $1$ and $100$ such that $f(c)=0$ by the intermediate value theorem.

Maple tells me there are in fact three solutions, and their analytic forms use the Lambert $W$ function, which I would guess you've never heard of (no implied insult to you, this is just a probabalistic thing). If you have heard of it, or are interested in it anyway, the values given are $$\frac{-2W(-\frac{\ln 2}{20})}{\ln 2}, \frac{-2W(-1,-\frac{\ln 2}{20})}{\ln 2}, \frac{-2W(\frac{\ln 2}{20})}{\ln 2}$$

The approximate values are $0.1036578164$, $14.32472784$, $-0.09670403432$.