How prove this inequality?? [closed]

How prove :

$${31}^{11}<{17}^{14}$$

closed as off-topic by user228113, A.P., Davide Giraudo, Thomas, Martin RDec 28 '15 at 14:33

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• With some extimations, I guess. I would not make the whole calculation, if I were you! – user228113 Dec 28 '15 at 13:39
• There is a big gap between the expressions, so this doesn't look too hard. – Element118 Dec 28 '15 at 13:39

I showed the numbers in diferent way:

${17}^{14}>{16}^{14}$

${16}^{14}={2^4}^{14}=2^{56}$

$17^{14}>2^{56}$

${31}^{11}<{32}^{11}$

$32^{11}={2^5}^{11}=2^{55}$

$2^{55}<2^{56}$

$17^{14}>2^{56}>2^{55}>31^{11}$

I think i show it more clear :)

$31^{11} < 32^{11} = 2^{55} < 2^{56} = 2^{4 \cdot 14} = 16^{14} < 17^{14}$

$$31^{11} < 32^{11} = (2^5)^{11}=2^{55} < 2^{56}=(2^4)^{14}=16^{14} < 17^{14}$$

$31^{11} < 34^{11}$

$34^{11} = (17\times2)^{11} = (17^{11}) \times (2^{11}) = 17^{11} \times 2^4 \times 2^4 \times 2^3$

$17^{11} \times 2^4 \times 2^4 \times 2^3 < 17^{11} \times 17 \times 17 \times 17$

$17^{11} \times 17 \times 17 \times 17 = 17^{14}$

Therefore, $31^{11} < 17^{14}$

• With mathjax it might have been clear. – user228113 Dec 28 '15 at 13:49
• @element118 thanks friend for valuable edit . – Tesla Dec 28 '15 at 13:51

Consider the : $$1 < \left(\frac{17}{31}\right)^{11}\cdot17^{3}$$ Now the major fact: $17^{3}>2^{11}$, i hope you could continue it!

$17^{14}=168377826559400929$ and $31^{11}=25408476896404831$. QED.

• I'm assuming this question is not to be done with a calculator. Even if you did this in your head/ on paper there are probably faster ways to do it. – Cataline Dec 28 '15 at 13:46
• @Cataline I know, I was just being a punk. And for the record I didn't use a calculator, I used command-line Python. – Gregory Grant Dec 28 '15 at 13:50
• I think that counts as a calculator... – Element118 Dec 28 '15 at 13:50
• @Element118 That's like calling Sam's club a corner convenience store. – Gregory Grant Dec 28 '15 at 13:56

Consider the following equivalent inequalities \begin{align} 31^{11} &< 17^{14} \\ 31^{10} \; 31 &< 17^{10} \; 17^4 \\ 31 \; \sqrt{31} &< 17 \; \sqrt{17^4} \end{align} and observe that the last one holds because $$\frac{31}{17} < 2 < \sqrt{\frac{17^3}{2}} < \sqrt{\frac{17^4}{31}}$$ since $17^3/2 > 17^2 \; 8$ and $17^2 > 12^2 = 144 > 128 = 2^7$.