# 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.)

Feb 8, 2018

I think both are equal.

#### Explanation:

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

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

Otherwise, I was thinking about logarithm concept.

Feb 9, 2018

See below.

#### Explanation:

Calling

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

both ${y}_{1} = {\left(x + \epsilon\right)}^{x}$ and ${y}_{2} = {x}^{x + \epsilon}$ are $> 1$

and $\ln \left(\cdot\right)$ is a strictly increasing function for $x \in {\mathbb{R}}^{+}$ hence if

${y}_{1} > {y}_{2} \Rightarrow \ln \left({y}_{1}\right) > \ln \left({y}_{2}\right)$ or
${y}_{1} < {y}_{2} \Rightarrow \ln \left({y}_{1}\right) < \ln \left({y}_{2}\right)$

now

$\ln \left({y}_{1}\right) - \ln \left({y}_{2}\right) = x \ln \left(x + \epsilon\right) - \left(x + \epsilon\right) \ln x = x \left(\ln \left(x + \epsilon\right) - \ln x\right) - \epsilon \ln x$

and consequently

$\frac{\ln \left({y}_{1}\right) - \ln \left({y}_{2}\right)}{\epsilon} = x \left(\frac{\ln \left(x + \epsilon\right) - \ln x}{\epsilon}\right) - \ln x$

but

$\left(\frac{\ln \left(x + \epsilon\right) - \ln x}{\epsilon}\right) \approx \frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$ hence

$\frac{\ln \left({y}_{1}\right) - \ln \left({y}_{2}\right)}{\epsilon} \approx x \left(\frac{1}{x}\right) - \ln x = 1 - \ln x$

Considering now that $\ln x \approx 0.881374$ we have

$\frac{\ln \left({y}_{1}\right) - \ln \left({y}_{2}\right)}{\epsilon} \approx 1 - 0.8813735870195429 > 0$ so concluding

$\ln \left({y}_{1}\right) > \ln \left({y}_{2}\right) \Rightarrow {\left(1 + \sqrt{2} + {10}^{- {10}^{6}}\right)}^{1 + \sqrt{2}} > {\left(1 + \sqrt{2}\right)}^{1 + \sqrt{2} + {10}^{- {10}^{6}}}$