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Inequality Problem

Prove that :

\(2^{135}+3^{133} < 4^ {108}\)

I don't know where this problem came,I'd be glad if you could cite the origin of the problem.And obviously please post any solution which you find.

Note by Soham Chanda
3 years, 7 months ago

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One-liner: \(2^{135} + 3^{133} = \left( \left(\tfrac{2}{3}\right)^{135} + \tfrac{1}{3^2}\right)3^{135} < 3^{135} = (3^5)^{27} = 243^{27} < 256^{27} = (4^4)^{27} = 4^{108}\). Jimmy Kariznov · 3 years, 7 months ago

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@Jimmy Kariznov Nice!

How about if we increase \(2^{135}\) to \(2^{215} \)? And how much higher can we 'easily' increase the power of 2? Calvin Lin Staff · 3 years, 7 months ago

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@Calvin Lin Since \(3^5 = 243 < 256 = 2^8\), we have \(2^{215} + 3^{133} < 2^{215} + 2^{212.8} < 2 \cdot 2^{215} = 2^{216} = 4^{108}\).

Also, \(2^{216} + 3^{133} > 2^{216} = 4^{108}\). So, the largest integer such that \(2^n + 3^{133} < 4^{108}\) is \(215\). Jimmy Kariznov · 3 years, 7 months ago

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@Jimmy Kariznov This is great!

A takeaway from this is that the order of magnitude is quite easy to estimate. Calvin Lin Staff · 3 years, 7 months ago

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Do you mean \(2^{135} + 2^{133} < 4^{108}\)?

If so, \(2^{135}+2^{133} = (2^2+1)2^{133} = 5 \cdot 2^{133} < 8 \cdot 2^{133} = 2^{136} < 2^{216} = 4^{108}\). Jimmy Kariznov · 3 years, 7 months ago

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@Jimmy Kariznov nono that was a typing mistake sorry Soham Chanda · 3 years, 7 months ago

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