Question

Log3(4x+8)-log3(x-3)=log3x

Answer 1

ОДЗ: 4x>-8; x>-2; x>0; x>3
x (3; +бесконечности)

(4x+8)/(x-3)=x. x не равен 3.
4x+8=x^2-3x
x^2-7x-8=0
D=49+32=81
x1=(7+9)/2= 8
x2= (7-9)/2 = -1 ( не подходит по одз)
Ответ: x=8

Answer 2

To solve the equation $\log_3(4x+8) - \log_3(x-3) = \log_3x$, we'll use the properties of logarithms. The following steps will lead us to the solution:

Step 1: Combine the logarithms on the left-hand side using the quotient rule of logarithms.

Step 2: Apply the power rule of logarithms to simplify the equation.

Step 3: Solve for $x$.

Let's proceed with the steps:

Step 1: Combine the logarithms on the left-hand side using the quotient rule.

$\log_3\left(\frac{4x+8}{x-3}\right) = \log_3x$

Step 2: Apply the power rule of logarithms.

Now, recall that $\log_b a = c$ can be rewritten as $b^c = a$.

So, we have:

$\frac{4x+8}{x-3} = x$

Step 3: Solve for $x$.

To solve the equation, we'll start by eliminating the fraction:

Multiply both sides by $(x - 3)$ to get rid of the denominator:

$(x-3) \cdot \frac{4x+8}{x-3} = x \cdot (x-3)$

Simplify:

$4x + 8 = x^2 - 3x$

Move all terms to one side to set the equation to zero:

$x^2 - 3x - 4x - 8 = 0$

Combine like terms:

$x^2 - 7x - 8 = 0$

Now, we have a quadratic equation. We can solve it by factoring:

$(x-8)(x+1) = 0$

Now, set each factor to zero and solve for $x$:

1. $x - 8 = 0$ -> $x = 8$
2. $x + 1 = 0$ -> $x = -1$

However, we must check the solutions in the original equation to ensure that they are valid.

Checking $x = 8$:

$\log_3(4(8)+8) - \log_3(8-3) = \log_3(8)$ $\log_3(32) - \log_3(5) = \log_3(8)$

Now, evaluate the logarithms:

$5 - \log_3(5) = \log_3(8)$

We can see that $5 - \log_3(5) \neq \log_3(8)$, so $x = 8$ is not a valid solution.

Checking $x = -1$:

$\log_3(4(-1)+8) - \log_3(-1-3) = \log_3(-1)$ $\log_3(4) - \log_3(-4) = \log_3(-1)$

Now, evaluate the logarithms:

$\log_3(4) - \text{(Not defined)} = \text{(Not defined)}$

Since $\log_3(-4)$ is not defined in the real number domain (logarithms of negative numbers are not real), $x = -1$ is also not a valid solution.

Therefore, the original equation has no valid solutions in the real number domain.