Question

Ordonati descrescator numerele ( cu tot cu rezolvare) : 25 ¹⁶ ; 2 ⁴⁸ ; 16 ²⁰;

Answer 1

$\hbox{Ridicarea unui numar la o putere:} \\ a^n=a\cdot a\cdot a\cdot ...\cdot a\cdot a\ \ \ \ \leftarrow\ de\ n\ ori\ a \\ \\ 1^m=1\ \forall\ m\in R\ (1\ la\ orice\ putere\ da\ 1) \\ a^0=1\ \forall\ a\in R\ (orice\ numar\ la\ puterea\ 0\ da\ 1) \\ 0^ \alpha =0\ \forall\ \alpha\in R\ (0\ la\ orice\ putere\ da\ 0) \\ 0^0\leftarrow\ operatie\ interzisa.\ \\ (alt\ exemplu\ de\ operatie\ interzisa\ este\ impartirea\ unui\ numar\ la\ 0)$

$Proprietati: \\ a^m\cdot a^n=a^{m+n} \\ a^m:a^n= \frac{a^m}{a^n} =a^{m-n}\ pentru\ a \neq 0 \\ (a^m)^n=a^{m\cdot n} \\ a^m\cdot b^m=(a\cdot b)^m \\ a^m:b^m= \frac{a^m}{b^m}= (\frac{a}{b})^m=(a:b)^m\ pentru\ b \neq 0$



$25^{16}=(5^2)^{16}=5^{32} \\ 16^{20}=(2^4)^{20}=2^{80} \\ 2^{16}=2^{16} \\ Avem\ 2^{16}\ \textless \ 2^{80} \\ Mai\ ramane\ de\ analizat\ cum\ este\ 5^{32}\ fata\ de\ aceste\ numere. \\ 2^{80}=2^{5\cdot2^4}=32^{2^4} \\ 25^{2^4}=25^{16} \\ 32^{2^4}=32^{16} \\ 25^{16}\ \textless \ 32^{16} \\ Deci:\ 25^{16}\ \textless \ 16^{20}$

$25^{16}\ ?\ 2^{48} \\ 2^{48}=2^{3\cdot16}=(2^3)^{16}=8^{16}\ \textless \ 25^{16} \\ \\ Deci:\ 2^{48}\ \textless \ 25^{16}\ \textless \ 16^{20}$