# What is the greater: 1000^(1000) or 1001^(999)?

Sep 6, 2016

${1000}^{1000} > {1001}^{999}$

#### Explanation:

Considering the equation

${1000}^{1000} = {1001}^{x}$

if $x > 999$

then

${1000}^{1000} > {1001}^{999}$

else

${1000}^{1000} < {1001}^{999}$

Appliying the log transformation to both sides.

$1000 \log 1000 = x \log 1001$

but

log 1001 = log1000+1/1000xx1-1/(2!)1/1000^2xx1^2+2/(3!)1/1000^3xx1^3 +cdots+1/(n!)(d/(dx)log x)_(x=1000)1^n.

This series is alternate and rapidly convergent so

$\log 1001 \approx \log 1000 + \frac{1}{1000}$

Substituting in

$x = 1000 \log \frac{1000}{\log 1000 + \frac{1}{1000}} = 1000 \left(\frac{3000}{3001}\right)$

but $\frac{3000}{3001} = 0.999667$ so

$x = 999.667 > 999$ then

${1000}^{1000} > {1001}^{999}$

Sep 6, 2016

Here's an alternative solution using the binomial theorem to prove:

${1001}^{999} < {1000}^{1000}$

#### Explanation:

By the binomial theorem:

(1+1/1000)^999 = 1/(0!) + 999/(1!)1/1000 + (999*998)/(2!)1/1000^2 + (999*998*997)/(3!) 1/1000^3 + ... + (999!)/(999!) 1/1000^999

< 1/(0!) + 1/(1!) + 1/(2!) + 1/(3!) +... = e ~~ 2.718

So:

${1001}^{999} = {\left(\frac{1001}{1000} \cdot 1000\right)}^{999}$

$\textcolor{w h i t e}{{1001}^{999}} = {\left(1 + \frac{1}{1000}\right)}^{999} \cdot {1000}^{999}$

$\textcolor{w h i t e}{{1001}^{999}} < e \cdot {1000}^{999} < 1000 \cdot {1000}^{999} = {1000}^{1000}$

Sep 7, 2016

${1000}^{1000} > {1001}^{999}$

#### Explanation:

#Use log 1000=log 10^3=3 and log 1001=3.0004340...

Here, the logarithms of the two are

$\log \left({1000}^{1000}\right) = 1000 \log 1000 = \left(1000\right) \left(3\right) = 3000$ and

$\log {1001}^{999} = \left(999\right) \left(3.0004340 \ldots\right) = 2997.4$...

As log is an increasing function,

${1000}^{1000} > {1001}^{999}$.