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Ecuatie logaritmica 1
Posted: Mon Dec 01, 2008 8:35 pm
by mihai++
Rezolvati in \( \mathbb{R}_+^* \):
\( x^{\log_2{3}}+x^{\log_3{2}}=2x. \)
Posted: Mon Dec 01, 2008 9:17 pm
by Marius Mainea
Daca \( x\ge 1 \), atunci
\( 2x=x^{\log_2 3}+x^{\log_3 2}\ge2\sqrt{x^{\log_2 3+\log_3 2}}\ge2x^{\sqrt{\log_2 3\cdot \log_3 2}}=2x, \)
de unde
\( x^{\log_2 3}=x^{\log_3 2}, \)
deci \( x=1 \).
Posted: Mon Dec 01, 2008 9:24 pm
by Laurian Filip
Asta mi-am dat seama si eu... Dar ce se intampla cand \( x<1 \)?
Posted: Mon Dec 01, 2008 9:32 pm
by Marius Mainea
Ne gandim.
Posted: Tue Dec 02, 2008 10:34 am
by mihai++
Eu am facut un grafic pe calculator (pt \( \sqrt{x}+x^2=2x \) am gasit o solutie intre 0.7 si 0.8 ) si cred ca ecuatia \( x^a+x^{\frac{1}{a}}=2x,\ a=\log_23 \), are o solutie subunitara care nu poate fi calculata.
Posted: Tue Dec 02, 2008 11:27 am
by Beniamin Bogosel
Exista o solutie in (0,1), \( x=0,3742... \). Nu cred ca se poate calcula.