perplexus.info :: Just Math : a KISS puzzle

a KISS puzzle (Posted on 2017-02-27)

Prove:

log_pie + log_epi >2

Looks complicated?
It is not!

See The Solution Submitted by Ady TZIDON

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Solution

| Comment 1 of 4

log_pie + log_epi = log_ee / log_epi + log_epi [change to common base e]

= 1/ln pi + ln pi . [usual notation for natural logs]

= 1/L + L , say

Now 1/L + L > 2 iff L^2 - 2L + 1 > 0 and L > 0

iff (L - 1)^2 > 0 and L > 0 ......... Eq. 1

Now L = ln pi > 1 since pi > e

Therefore, Eq. 1 is true and the result is proved.

Nice problem, Ady!

Edited on February 27, 2017, 9:58 am

Posted by JayDeeKay on 2017-02-27 09:52:04

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