Question

Log3(18-x^2)-log3(4-x)=log3(2x+3)

Answer 1

log₃(18-x²)÷(4-x)= log₃(2x+3)

18-x²÷ 4-x = 2x+3

18-x²÷4-x=((2x+3)(4-x))÷4-x

18-x²+2x²-5x-12=0

x²-5x+6=0

Д = 25-24=1

x₁ = 3

x₂ = 2

Answer 2

3 и 2

Пошаговое объяснение:

log₃(18-x²)-log₃(4-x)=log₃(2x+3)

$\frac{log(18-x^2)}{log_3} -\frac{log(4-x)}{log_3} =\frac{log(2x+3)}{log_3}$

$\frac{log(18-x^2)-log(4-x)}{log_3} =\frac{log(2x+3)}{log_3}$

$log(18-x^2)-log(4-x)=log(2x+3)$

$-log(4-x)-log(2x+3)+log(18-x^2)=0$

$log(\frac{18-x^2}{(4-x)(2x+3)}) =1$

$18-x^2=(4-x)(2x+3)$

$18-x^2=-2x^2+5x+12$

$x^2-5x+6=0$

$(x-3)(x-2)=0$

$x=3$

$x-2=0$

$x=2$

Удачи!