Question

Comparati:
2 la puterea 302 si 3 la puterea 203 .

Answer 1

Hmm, presupunem ca $2^{302}\ \textless \ 3^{203}$. Ca sa verificam, logaritmam expresia si vedem ce da:
$2^{302}\ \textless \ 3^{203} \Leftrightarrow \lg2^{302}\ \textless \ \lg3^{203}\Leftrightarrow 302\lg2\ \textless \ 203\lg3 \Leftrightarrow \frac{302}{203} \ \textless \ \frac{\lg3}{\lg2} \\ \Leftrightarrow \frac{302}{203} \ \textless \ \log_2 3.$
Dar $\log_23 = \log_2 \sqrt{9} \ \textgreater \ \log_2 \sqrt{8} = \log_2 2^{\frac{3}{2}} = \frac{3}{2} \Rightarrow \log_23 \ \textgreater \ \frac{3}{2}.$
$\frac{3}{2} \ \textgreater \ \frac{302}{203}$ (se verifica usor prin calcul).
$\Rightarrow \log_23 \ \textgreater \ \frac{302}{203}.$ Deci presupunerea facuta este adevarata.
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Si pentru clasa a 5-a: :D
[tex]2^{302} = 2^2 \cdot 2^{300} = 4 \cdot (2^3)^{100} = 4 \cdot 8^{100}.\\ 3^{203} = 3^3 \cdot 3^{200} = 27 \cdot (3^2)^{100} = 27 \cdot 9^{100}[/tex], de unde se vede clar ca $2^{302}\ \textless \ 3^{203}$.