How to prove $A = \log 201 - \frac{201}{200}*\log 2 - 2 > 0$ without using a calculator?

Question

I found a problem online about comparing 2 numbers. I tried doing it the logarithm way and reached this point, but I'm not sure if it's possible to prove that A > 0 without using a calculator.

EDIT: Log has base 10.

Please advise. Thank you!

Answer 1

\begin{align*} A &=\log 201 - \frac{201}{200}\log 2 - 2\\ &=\log 201 - \frac{201}{200}\log 2 - \log 100\\ &=\log \frac{201}{100}-\frac12 \frac{201}{100}\log 2\\ \end{align*}

Now define $f(x)=\log x-\frac x2\log 2$. You can easily check that $f''(x)$ is negative on its domain (so $f$ is concave). Now since $f(2)=f(4)=0$, the continuous $f(x)$ must be positive for $2<x<4$.