The given equation is:

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

Step 1: Combine the logarithmic terms on the left-hand side

Using the logarithmic property $\ln(a) + \ln(b) = \ln(a \cdot b)$, we rewrite the left-hand side:

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

Step 2: Equating the arguments of the logarithms

Since the natural logarithmic function $\ln(x)$ is one-to-one, we can equate the arguments:

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

Step 3: Expand the left-hand side

Expand the product on the left:

$2x^2 + 3x - 6x - 9 = -4x^2$

Simplify:

$2x^2 - 3x - 9 = -4x^2$

Step 4: Combine like terms

Bring all terms to one side of the equation:

$2x^2 + 4x^2 - 3x - 9 = 0$

$6x^2 - 3x - 9 = 0$

Step 5: Simplify the equation

Factor out a 3:

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

Step 6: Solve the quadratic equation

Focus on solving $2x^2 - x - 3 = 0$. Use the quadratic formula:

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

Here, $a = 2$, $b = -1$, and $c = -3$. Substitute into the formula:

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

$x = \frac{1 \pm \sqrt{1 + 24}}{4}$

$x = \frac{1 \pm \sqrt{25}}{4}$

$x = \frac{1 \pm 5}{4}$

Step 7: Compute the roots

$x = \frac{1 + 5}{4} = \frac{6}{4} = \frac{3}{2}$

$x = \frac{1 - 5}{4} = \frac{-4}{4} = -1$

Step 8: Check for extraneous solutions

The arguments of the logarithmic functions must be positive. Substituting $x = -1$ into the original equation:

$\ln(x-3) = \ln(-4), \quad \ln(2x+3) = \ln(1),$

which is invalid since logarithms of negative numbers are undefined. Therefore, $x = -1$ is not a solution.

Substituting $x = \frac{3}{2}$:

$x - 3 = \frac{3}{2} - 3 = -\frac{3}{2}, \quad \ln(-\frac{3}{2})$

is also invalid. Hence, no solution satisfies the equation.

Final Answer:

The equation has no solution.

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Would you like further clarification or details? Here are some related questions:

1. What are the rules for solving logarithmic equations?
2. Why can't logarithms of negative numbers be calculated in real numbers?
3. Can this equation have solutions in the complex domain?
4. What are some examples of logarithmic equations with valid solutions?
5. How do the properties of logarithms simplify solving equations?

Tip: Always check for restrictions (domain) when solving logarithmic equations to avoid extraneous solutions.