How can I compose the proof?

Question

Prove that $$\Big(\frac{21}{20}\Big)^{100}>100.$$ I have tried proving that $$100\log\Big(\frac{21}{20}\Big)>2$$ but I was not able to evaluate it properly.

Answer 1

I think you are on the right track. Using the basic rules for logartihms we get:

$\log(\frac{21^{100}}{20}) = 100\log(21) - \log(20) > 100 - 1 - \log(2) > 2$

Answer 2

\begin{align}\Big({21\over20}\Big)^{100} &= (1+{1\over 20})^{100}\\ &={100\choose 0} +{100\choose 1} ({1\over 20})^{1} +{100\choose 2} ({1\over 20})^{2} +{100\choose 3} ({1\over 20})^{3} +... \\ &>1+5+12+20+24,5+23,5+18 =105 \end{align}