Question

$log_{5} \frac{4x^{2}-1 }{3-x}+ log_{5} \frac{3-x}{9+3x}=0$

Answer 1

$\frac{4 x^{2} -1}{3-x}* \frac{3-x}{9+3x}=1 \\ \frac{4 x^{2} -1}{9+3x}=1 \\ x \neq -3 \\ 4 x^{2} -1=9+3x \\ 4 x^{2} -3x-10=0$
Δ=169
$x_{1}= \frac{3-13}{8}=- \frac{5}{4} \\ x_{2}= \frac{3+13}{8}=2$
S={2}

Answer 2

㏒₅[(4x²-1)/(3-x)·(3-x)/(9+3x)]=0
se simplifica 3-x cu 3-x si obtinem
(4x²-1)/(9+3x)=1
4x²-1=9+3x
4x²-3x-10=0
x₁=2
x₂=-5/4