×

# 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

## Comments

Sort by:

Top Newest

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

Log in to reply

Nice!

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

Log in to reply

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 $$215$$. · 3 years, 7 months ago

Log in to reply

This is great!

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

Log in to reply

Do you mean $$2^{135} + 2^{133} < 4^{108}$$?

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

Log in to reply

nono that was a typing mistake sorry · 3 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...