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Question Number 21781 by Tinkutara last updated on 03/Oct/17

Which is greater 10^(11) or 11^(10) ?

$$\mathrm{Which}\:\mathrm{is}\:\mathrm{greater}\:\mathrm{10}^{\mathrm{11}} \:\mathrm{or}\:\mathrm{11}^{\mathrm{10}} ? \\ $$

Answered by Joel577 last updated on 03/Oct/17

10^(11) ... 11^(10) 11 log 10 ... 10 log 11 11 ... 10 (1.04) 11 > 10.4

$$\:\:\:\:\:\:\:\:\:\:\mathrm{10}^{\mathrm{11}} \:...\:\mathrm{11}^{\mathrm{10}} \\ $$$$\mathrm{11}\:\mathrm{log}\:\mathrm{10}\:...\:\mathrm{10}\:\mathrm{log}\:\mathrm{11} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{11}\:\:...\:\mathrm{10}\:\left(\mathrm{1}.\mathrm{04}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{11}\:\:>\:\mathrm{10}.\mathrm{4} \\ $$

Commented by Tinkutara last updated on 03/Oct/17

But without calculator how can be done?

$$\mathrm{But}\:\mathrm{without}\:\mathrm{calculator}\:\mathrm{how}\:\mathrm{can}\:\mathrm{be}\:\mathrm{done}? \\ $$

Answered by Bruce Lee last updated on 04/Oct/17

We have this form (n+1)^n <n^(n+1) ;n≥3 proof: for n=3 ⇒ 4^3 <3^4 or 64<81 true assum that it true till n=k we get (k+1)^k <k^(k+1) we wish to prove that (k+2)^(k+1) <(k+1)^(k+2ww) k^(k+1) >(k+1)^k ⇒k^(k+1) ∙(((k+1)^(k+2) )/k^(k+1) )>(k+1)^k ∙(((k+1)^(k+2) )/k^(k+1) ) (k+1)^(k+2) >(((k+1)^(2k+2) )/k^(k+1) ) (k+1)^(k+2) >(((k^2 +2k+1)^(k+1) )/k^(k+1) ) (k+1)^(k+2) >(k+2+(1/k))^(k+1) >(k+2)^(k+1) ⇒(k+2)^(k+1) <(k+1)^(k+2) ⇒(n+1)^n <n^(n+1)

$$\boldsymbol{\mathrm{We}}\:\boldsymbol{\mathrm{have}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{form}}\:\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} <\boldsymbol{\mathrm{n}}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:;\boldsymbol{\mathrm{n}}\geqslant\mathrm{3} \\ $$$$\boldsymbol{\mathrm{proof}}: \\ $$$$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{n}}=\mathrm{3}\:\Rightarrow\:\mathrm{4}^{\mathrm{3}} <\mathrm{3}^{\mathrm{4}} \:{or}\:\mathrm{64}<\mathrm{81}\:\:\:\boldsymbol{\mathrm{true}} \\ $$$$\boldsymbol{\mathrm{assum}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{true}}\:\boldsymbol{\mathrm{till}}\:\boldsymbol{\mathrm{n}}=\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{get}}\:\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} <\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{wish}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\left(\boldsymbol{\mathrm{k}}+\mathrm{2}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} <\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}\boldsymbol{\mathrm{ww}}} \\ $$$$\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} >\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \:\:\: \\ $$$$\Rightarrow\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} \centerdot\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} }{\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} }>\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \centerdot\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} }{\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} >\frac{\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{2}} }{\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} >\frac{\left(\boldsymbol{\mathrm{k}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} }{\boldsymbol{\mathrm{k}}^{\boldsymbol{\mathrm{k}}+\mathrm{1}} } \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} >\left(\boldsymbol{\mathrm{k}}+\mathrm{2}+\frac{\mathrm{1}}{\boldsymbol{\mathrm{k}}}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} >\left(\boldsymbol{\mathrm{k}}+\mathrm{2}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} \\ $$$$\Rightarrow\left(\boldsymbol{\mathrm{k}}+\mathrm{2}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{1}} <\left(\boldsymbol{\mathrm{k}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}+\mathrm{2}} \\ $$$$\Rightarrow\left(\boldsymbol{\mathrm{n}}+\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} <\boldsymbol{\mathrm{n}}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 04/Oct/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Commented by Joel577 last updated on 04/Oct/17

nice proof Sir

$${nice}\:{proof}\:{Sir} \\ $$

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