text: 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.
  
  $4^{80}=2^{160}$
  
  $2^{160}-2^{100}=2^{100}(2^{60}-1)$
  
  $ln 2^{100}(2^{60}-1)=ln 2^{100}+ln (2^{60}-1)=100 ln 2+41.59=69.31+41.59=110.9$
  
  $ln 3^{100}=100ln 3=109.86$
  
  Thus, $ln 3^{100}
  
  $3^{100}<2^{100}(2^{60}-1)$
  
  $2^{100}+3^{100}<2^{160}=4^{80}$
  
  3. Originally Posted by alexmahone
  I can't think of an elegant way of doing this.
  
  $4^{80}=2^{160}$
  
  $2^{160}-2^{100}=2^{100}(2^{60}-1)$
  
  $ln 2^{100}(2^{60}-1)=ln 2^{100}+ln (2^{60}-1)=100 ln 2+41.59=69.31+41.59=110.9$
  
  $ln 3^{100}=100ln 3=109.86$
  
  Thus, $ln 3^{100}
  
  $3^{100}<2^{100}(2^{60}-1)$
  
  $2^{100}+3^{100}<2^{160}=4^{80}$
  ...and without unnecessary calculation...?
  
  4. thnks alots..
  
  5. Mine is not so beautiful but you can take a look :
  
  Consider
  
  $2^8 = 256 = 243 + 13 = 3^5 + 13$ so we have
  
  $2^{160} = (3^5 + 13)^{20} > 3^{100} + 20(3^{95})(13)$
  
  $= 3^{100} + (260)3^{95} > 3^{100} + (243)3^{95} = 3^{100} + 3^{100} > 3^{100} + 2^{100}$
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