Question

Sa se rezolve:
$log_{3x}( \frac{3}{x}) + log^2_{3} x=1$

Answer 1

Facem transformarea bazei: $log_{3x} \frac{3}{x}= \frac{ log_{3} \frac{3}{x} }{ log_{3}3x }= \frac{1- log_{3}x }{ 1+log_{3}x }$, notam: $log_{3}x=y$, si obtinem ecuatia: $\frac{1-y}{1+y}=(1-y)(1+y),sau,(1-y)-(1-y)(1+y)^2=0,sau,$
$(1-y)[1-(1+y)^2]=0,deci, y_{1}=1,adica, log_{3}x=1,de,unde, x_{1} =3$. A doua paranteza se descompune in produs de suma prin diferenta: $(1-1-y)(1+1+y)=0,rezulta, y_{2}=0,sau, log_{3}x=0,deci, x_{2}=1,$ si 2+y=0⇒$y_{3}=-2,sau, log_{3}x=-2,deci, x_{3}= 3^{-2}= \frac{1}{9}$