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

Ответ:

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!

Answer 3

To solve the given equation for $x$:

$\log_3(18 - x^2) - \log_3(4 - x) = \log_3(2x + 3)$

We can use logarithmic properties to simplify the equation. The difference of logarithms can be combined into a single logarithm with a division:

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

Now we have a single logarithmic equation with the same base. In order for the logarithms to be equal, their arguments must be equal as well:

$\frac{18 - x^2}{4 - x} = 2x + 3$

Now let's solve for $x$:

Cross-multiply to eliminate the fraction:

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

Expand the right side:

$18 - x^2 = 8x - 2x^2 + 12 - 3x$

Combine like terms:

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

Divide both sides by 2:

$0 = 3x - x^2 - 3$

Rearrange the terms:

$x^2 - 3x + 3 = 0$

This is a quadratic equation. To solve it, you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -3$, and $c = 3$. Plug these values into the formula:

$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

$x = \frac{3 \pm \sqrt{9 - 12}}{2}$

$x = \frac{3 \pm \sqrt{-3}}{2}$

Since the square root of a negative number is not a real number, there are no real solutions for $x$ that satisfy the given equation. The solution set is empty, meaning there are no values of $x$ that satisfy the original equation.