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.

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.

No vote yet

7 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestOne-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, 5 months ago

Log in to reply

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, 5 months ago

Log in to reply

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, 5 months ago

Log in to reply

A takeaway from this is that the order of magnitude is quite easy to estimate. – Calvin Lin Staff · 3 years, 5 months ago

Log in to reply

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, 5 months ago

Log in to reply

– Soham Chanda · 3 years, 5 months ago

nono that was a typing mistake sorryLog in to reply