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

3 Answers
Sep 6, 2016

Answer:

#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

#log1001 approx log1000+1/1000#

Substituting in

#x = 1000 log1000/(log1000+1/1000)=1000(3000/3001)#

but #3000/3001 = 0.999667# so

#x = 999.667 > 999# then

#1000^1000 > 1001^999#

Sep 6, 2016

Answer:

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 = (1001/1000 * 1000) ^ 999#

#color(white)(1001^999) = (1+1/1000)^999 * 1000^999#

#color(white)(1001^999) < e*1000^999 < 1000*1000^999 = 1000^1000#

Sep 7, 2016

Answer:

#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( 1000 ^ 1000)=1000 log1000= (1000)(3) = 3000# and

#log 1001^999= (999)(3.0004340...)=2997.4#...

As log is an increasing function,

#1000^1000 > 1001^999#.