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 !! :)

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Answer 1 · 2018-02-08T05:18:31

I think both are equal.

#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.

Answer 2 · 2018-02-09T14:31:34

See below.

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))#