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cum se rezolva : log baza 3 din x -log baza 3 (x+8) = -log baza 3( x+3) ? va rog ajutati-ma


Răspuns :

log3 x - log3 (x+8) = -log3 (x+3)
log3 x + log3 (x+3) = log3 (x+8)
log3 [x(x+3)] = log3 (x+8)
x(x+3)=x+8
x^2+3x=x+8
x^2+2x-8=0
(x-2)(x+4)=0
x=2; x=-4 dar x>0, atunci S: x=2
㏒₃x-㏒₃(x+8)=-㏒₃(x+3)⇒
avem aceasi baza 
CE X+8≥0⇒X≥-8
X+3≥0⇒X≥-3
X-X-8=X+3⇒-8-3=X⇒-11
SUCEES