Question

lg(2x+6)-2lg(2x-3)=1

Answer 1

DVA:x∈(3/2;∞)
lg(2x+6)=lg10+2lg(2x-3)
$lg(2x+6)=lg10+lg(2x-3)^2 \\ 2x+6=10(2x-3)^2 \\ 2x+6=10(4 x^{2} -12x+9) \\ 40 x^{2} -122x+84=0/:2 \\ 20 x^{2} -61x+42=0$
Δ=361
$x_{1}= \frac{61-19}{40}= \frac{42}{40}= \frac{21}{20} \\ x_{2}= \frac{61+19}{40}=2$
S={2}

Answer 2

lg(2x+6)-lg(2x-3)²=1
lg(2x+6)/(2x-3)²=1
(2x+6)/(2x-3)²=10
2x+6=10(4x²-12x+9)
40x²-122x+84=0
20x²-61x+42=0
x₁=1
x₂=21/40