Ecuatie logaritmica 1

mihai++ · Post by **mihai++** » Mon Dec 01, 2008 8:35 pm

Rezolvati in \( \mathbb{R}_+^* \):
\( x^{\log_2{3}}+x^{\log_3{2}}=2x. \)

Marius Mainea · Post by **Marius Mainea** » Mon Dec 01, 2008 9:17 pm

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 \( x=1 \).

Post by **Laurian Filip** » Mon Dec 01, 2008 9:24 pm

Asta mi-am dat seama si eu... Dar ce se intampla cand \( x<1 \)?

Marius Mainea · Post by **Marius Mainea** » Mon Dec 01, 2008 9:32 pm

Ne gandim.

mihai++ · Post by **mihai++** » Tue Dec 02, 2008 10:34 am

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.

Post by **Beniamin Bogosel** » Tue Dec 02, 2008 11:27 am

Exista o solutie in (0,1), \( x=0,3742... \). Nu cred ca se poate calcula.