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.

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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 - 7 years, 10 months ago

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 - 7 years, 10 months ago

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 - 7 years, 10 months ago

@Jimmy Kariznov – This is great!

A takeaway from this is that the order of magnitude is quite easy to estimate.

Calvin Lin Staff - 7 years, 10 months ago

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 - 7 years, 10 months ago

nono that was a typing mistake sorry

Soham Chanda - 7 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$