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

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

(I have hints available, if necessary.)

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Answer 1 · 2018-02-08T05:18:31

I think both are equal.

Explanation:

$10^{- 9000} \to 0$ $10^-9000->0$ as its value is negligible.

Then ${(1 + \sqrt{2})}^{1 + \sqrt{2} + 0} = {(1 + \sqrt{2} + 0)}^{1 + \sqrt{2}}$ $(1+sqrt2)^(1+sqrt2+0)=(1+sqrt2+0)^(1+sqrt2)$ .

Otherwise, I was thinking about logarithm concept.

Answer 2 · 2018-02-09T14:31:34

See below.

Explanation:

Calling

$x = 1 + \sqrt{2}$ $x = 1+ sqrt2$
$ε = 10^{- 10^{6}}$ $epsilon=10^(-10^6)$ we have

both $y_{1} = {(x + ε)}^{x}$ $y_1=(x+epsilon)^x$ and $y_{2} = x^{x + ε}$ $y_2=x^(x + epsilon)$ are $> 1$ $> 1$

and $ln (\cdot)$ $ln(cdot)$ 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

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

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

(I have hints available, if necessary.)

2 Answers

Explanation:

Explanation:

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