Question

Cum demonstrez ca 5 la puterea 2015 este mai mic decat 2 la puterea 4705?

Answer 1

Ai rezolvarea in poza de mai jos.

Answer 2

Se calculeaza cmmdc(2015,4705)
Pentru aceasta se descompune numerele 2015 si 4705 in factori primi
2015/5
403/31
13/13
1/
2015=5^1*31^1*13^1
4705/5
941/941
1/
4705=5^1*941^1
Se iau toti factorii primi comuni, la puterile cele mai mici
cmmdc(2015,4705)=5^1=5
[tex]5^{2015}=5^{403*5}=(5^{403}})^{5}\\2^{4705}=2^{941*5}=(2^{941})^{5}\\A\ compara\ (5^{403})^{5}\ cu\ (2^{941})^{5}\ este\ echivalent\ cu\ a\ compara\\ 5^{403}\ cu\ 2^{941}\\2^{5}\ \textgreater \ 5^{2}\\32\ \textgreater \ 25\\Deci:\\ 2^{941}=2^{5*188,2}=(2^{5})^{188,2}=32^{188,2}\\ 5^{403}=5^{2*201,5}=(5^{2})^{201,5}=25^{201,5}\\ \frac{32}{25}=\frac{32*4}{25*4}=\frac{128}{100}=1,28\\Astfel:\\ 32^{188,2}=(25*1,28)^{188,2}=25^{1,28*188,2}=25^{240,896}\ \textgreater \ 25^{201,5}\\ Astfel:\\32^{188,2}\ \textgreater \ 25^{201,5}\\2^{941}\ \textgreater \ 5^{403}\\5^{2015}\ \textless \ 2^{4705}[/tex]