2
$\begingroup$

I want to find out if $\frac{2^{256}}{1.158*10^{77}}$ is

a) $< 1$

b) $= 1$

c) $> 1$

What I tried so far

Thanks to the help of some comments that suggested to use $log$:

Upper bound for the top

With $log_{10} 2 = 0.30102... < 0.30103$ I can show that the top has an upper bound at

$256*log_{10} 2 < 256 * 0.30103 = 77.06368$

Lower bound for the bottom

With $log_{10} 1.158 = 0.063708... > 0.0637$ I can show that the bottom has a lower bound at

$log_{10} (1.158*10^{77})$ = $log_{10} (1.158)$ + $77$ $>$ $0.0637+77 = 77.0637$

Lets compare the bounds

$\frac{77.06368}{77.0637}$ so this would result in $< 1$

Is this enough?

  • 0
    Are you looking for a way without a calculator ? If not, just logarithm the expression. – 2017-02-13
  • 0
    The expression is close to $1$, a bit smaller. It will be hard to prove it without a calculator. – 2017-02-13
  • 0
    Yes a way without calculator, thought I can simplify it a bit more so that it directly can be seen. – 2017-02-13
  • 0
    Try the expansion of $\ln(1+x)$, but I think we need many terms to decide the magnitude. – 2017-02-13

1 Answers 1

3

$$2^{256}=10^{\log_{10}{2^{256}}}=10^{256\log_{10}2}$$ $$\begin{align}\frac{2^{256}}{1.158\times10^{77}}&=\frac{10^{256\log_{10}2}}{1.158\cdot10^{77}}\\&=\frac{10^{256\log_{10}2-77}}{1.158}\\ &=\frac{10^{256\log_{10}2-77}}{10^{\log_{10}1.158}}\\ &=10^{256\log_{10}2-77-\log_{10}1.158}\end{align}$$

By evaluating, $$256\log_{10}2-77-\log_{10}1.158\lt0$$ So, $$\frac{2^{256}}{1.158\times10^{77}}\lt1$$