Comparison between two numbers

[anemone]

Compare the numbers $2^{2016}$ and $3^{201}7^{604}$.

Spoiler

I don't have the time yet to try it, but I can tell this is a very delicious problem so I decided to make it as a challenge here and hopefully I can crack it when I've the time and am able. I hope too that you'll agree with me that this is a superb challenging problem and I'm looking forward to see how our members are going to solve it.

 

[kaliprasad]

Compare the numbers $2^{2016}$ and $3^{201}7^{604}$.

Spoiler

I don't have the time yet to try it, but I can tell this is a very delicious problem so I decided to make it as a challenge here and hopefully I can crack it when I've the time and am able. I hope too that you'll agree with me that this is a superb challenging problem and I'm looking forward to see how our members are going to solve it.

Spoiler

we have $3 * 7^3 = 1029 > 2^{10}$
so $\frac {3 * 7^3}{2^{10}} = \frac{1029}{1024} = 1 + \frac{5}{1024} < 1 + \frac{1}{201}$
hence $ (\frac {3 * 7^3}{2^{10}})^{201} < (1 + \frac{1}{201})^{201} < e $ as $(1+\frac{1}{x})^x < e$
so $ (3 * 7 ^3)^{201} < e * 2^{2010}$
or $3^{201} * 7^{603} < 3 * 2^{2010}$
or $3^{201} * 7^{604} < 21 * 2^{2010} < 64 * 2^{2010}$
or $3^{201} * 7^{604} < 2^{2016}$
hence $2^{2016}$ is larger

 

[anemone]

Spoiler

we have $3 * 7^3 = 1029 > 2^{10}$
so $\frac {3 * 7^3}{2^{10}} = \frac{1029}{1024} = 1 + \frac{5}{1024} < 1 + \frac{1}{201}$
hence $ (\frac {3 * 7^3}{2^{10}})^{201} < (1 + \frac{1}{201})^{201} < e $ as $(1+\frac{1}{x})^x < e$
so $ (3 * 7 ^3)^{201} < e * 2^{2010}$
or $3^{201} * 7^{603} < 3 * 2^{2010}$
or $3^{201} * 7^{604} < 21 * 2^{2010} < 64 * 2^{2010}$
or $3^{201} * 7^{604} < 2^{2016}$
hence $2^{2016}$ is larger

Very well done kaliprasad!(Cool)