1. ## diopantine equation

Which one is greater:2^100+3^100 or 4^80?

2. Originally Posted by dila
Which one is greater:2^100+3^100 or 4^80?
I can't think of an elegant way of doing this.

$\displaystyle 4^{80}=2^{160}$

$\displaystyle 2^{160}-2^{100}=2^{100}(2^{60}-1)$

$\displaystyle ln 2^{100}(2^{60}-1)=ln 2^{100}+ln (2^{60}-1)=100 ln 2+41.59=69.31+41.59=110.9$

$\displaystyle ln 3^{100}=100ln 3=109.86$

Thus, $\displaystyle ln 3^{100}<ln 2^{100}(2^{60}-1)$

$\displaystyle 3^{100}<2^{100}(2^{60}-1)$

$\displaystyle 2^{100}+3^{100}<2^{160}=4^{80}$

3. Originally Posted by alexmahone
I can't think of an elegant way of doing this.

$\displaystyle 4^{80}=2^{160}$

$\displaystyle 2^{160}-2^{100}=2^{100}(2^{60}-1)$

$\displaystyle ln 2^{100}(2^{60}-1)=ln 2^{100}+ln (2^{60}-1)=100 ln 2+41.59=69.31+41.59=110.9$

$\displaystyle ln 3^{100}=100ln 3=109.86$

Thus, $\displaystyle ln 3^{100}<ln 2^{100}(2^{60}-1)$

$\displaystyle 3^{100}<2^{100}(2^{60}-1)$

$\displaystyle 2^{100}+3^{100}<2^{160}=4^{80}$
...and without unnecessary calculation...?

4. thnks alots..

5. Mine is not so beautiful but you can take a look :

Consider

$\displaystyle 2^8 = 256 = 243 + 13 = 3^5 + 13$ so we have

$\displaystyle 2^{160} = (3^5 + 13)^{20} > 3^{100} + 20(3^{95})(13)$

$\displaystyle = 3^{100} + (260)3^{95} > 3^{100} + (243)3^{95} = 3^{100} + 3^{100} > 3^{100} + 2^{100}$