7

Put $2^{600}$, $3^{500}$, $4^{400}$, $5^{300}$, and $6^{200}$ in order.

Problem I found while looking at old problems from math competitions.

Clearly a simple solution would be to compare $600\ln2$, $500\ln3$, etc. But how would one go about solving this problem without a calculator? Expressing the numbers in the same base somehow?

Would appreciate any insight.

deadgrips
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  • Obviously you can divide all of those by 100 without changing their order: $6\log2$, $5\log3$, $4\log4$, $3\log5$, $2\log6$. And these, well -- you can convert those back: $2^6$, $3^5$... – Dan Uznanski Nov 06 '14 at 03:58
  • @DanUznanski Evaluating log 2, log 3 etc still rely on the calculator. – Mick Nov 06 '14 at 04:21
  • @DanUznanski Pls ignore my previous comment. I have it deleted but it still shows. – Mick Nov 06 '14 at 04:27
  • I deleted my answer because the author now specified an order. – David Nov 15 '14 at 22:18

3 Answers3

16

As $(a^b)^{100} = a^{100b}$ for all positive $a$ and $b$, this problem is the same as ordering

$2^6, 3^5, 4^4, 5^3$ and $6^2$

which is straight forward.

Simon S
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6

Surely you can arrange $2^6$, $3^5$, $4^4$, and so on in order without a calculator.

André Nicolas
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2

To evaluate these values, they must be placed on the same “platform”.

Re-writing $2^{600}$ as $(2^6)^{100}$; and $3^{500}$ as $(3^5)^{100}$ and so on will make them comparable (now), under the same platform.

After doing that, then, you are comparing $(64)^{100}, (243)^{100}$ and so on.

Mick
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