Solve for $N$: $2000N=(0.9025)^{\log_2 N}$

Question

I want to find the value of $N$ while $2000N=(0.9025)^{\log_2 N}$ (This is sample value not actual)

How to solve it?

The Whole Question which i am solving is $Pe=(Pt/N)(1-δ)^{\log_2 N}$

Where given values are If we use δ = 0.05, Pt = 1 mW, and Pe = 0.1 μW

Answer 1

We'll take $\log_2$ of both sides. So,

$$2000N=0.9025^{\log_2N}$$ $$\log_2 2000N=\log_2 0.9025^{\log_2 N}$$ $$\log_2 2000 + \log_2 N=(\log_2 N)(\log_2 0.9025)$$ $$\log_2 2000=(\log_2 N)(\log_2 0.9025)-\log_2 N$$ $$\log_2 2000=(\log_2 N)((\log_2 0.9025)-1)$$ $$\frac{\log_2 2000}{(\log_2 0.9025)-1}=\log_2 N$$ $$N=2^{\frac{\log_2 2000}{(\log_2 0.9025)-1}}$$

Or you can write: $$N=2000^{\left((\log_2 0.9025)-1\right)^{-1}}$$