Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ 
Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.

I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 \cdot 1976^7>0$. How should I do that?

Seems really simple - am I missing something?
$a = 4\times 1976^3 < 4\times 2000^3 = 32\times 10^9$
$\implies 1/a > \frac{1}{32}10^{-9} > 3\times 10^{-11}$
$b= 16\times 1976^7 > 10^{22}$ and $1/b <10^{-22}$
$\implies \frac{1}{a} - \frac{1}{b} > 3 \times 10^{-11} - 10^{-22} >10^{-19.76}$  (howsoever you interpret that exponent)
