Which is bigger: ( 1 + \sqrt{2} )^{ 1 + \sqrt{2} + 10^{-9,000} } or ( 1 + \sqrt{2} + 10^{-9,000} )^{ 1 + \sqrt{2} } ? If your calculator could actually handle this -- please put it away !! :)

(I have hints available, if necessary.)

2 Answers
Feb 8, 2018

I think both are equal.

Explanation:

10^-9000->0 as its value is negligible.

Then (1+sqrt2)^(1+sqrt2+0)=(1+sqrt2+0)^(1+sqrt2).

Otherwise, I was thinking about logarithm concept.

Feb 9, 2018

See below.

Explanation:

Calling

x = 1+ sqrt2
epsilon=10^(-10^6) we have

both y_1=(x+epsilon)^x and y_2=x^(x + epsilon) are > 1

and ln(cdot) is a strictly increasing function for x in RR^+ hence if

y_1 > y_2 rArr ln(y_1) > ln(y_2) or
y_1 < y_2 rArr ln(y_1) < ln(y_2)

now

ln(y_1)-ln(y_2) = x ln(x+epsilon) - (x+epsilon)ln x = x(ln(x+epsilon)-lnx) -epsilon lnx

and consequently

(ln(y_1)-ln(y_2))/epsilon = x((ln(x+epsilon)-lnx)/epsilon)-lnx

but

((ln(x+epsilon)-lnx)/epsilon) approx d/(dx)lnx = 1/x hence

(ln(y_1)-ln(y_2))/epsilon approx x(1/x)-lnx = 1-lnx

Considering now that lnx approx 0.881374 we have

(ln(y_1)-ln(y_2))/epsilon approx 1- 0.8813735870195429 > 0 so concluding

ln(y_1) > ln(y_2) rArr (1+sqrt2+10^(-10^6))^(1+sqrt2) > (1+sqrt2)^(1+sqrt2+10^(-10^6))