Solve the Log Equation: 2log(x+1) = log(2x²+8x)

To solve this problem, we'll follow these steps:

• Step 1: Apply the power rule of logarithms.
• Step 2: Formulate a quadratic equation.
• Step 3: Solve the quadratic equation.
• Step 4: Verify the solution is within the domain of the original logarithmic functions.

Now, let's work through each step:
Step 1: The equation is given by $2\log(x+1) = \log(2x^2 + 8x)$. By applying the power rule, $2\log(x+1)$ becomes $\log((x+1)^2)$. Hence, the equation becomes:

$\log((x+1)^2) = \log(2x^2 + 8x)$

Step 2: Since the logarithms are equal, we can equate their arguments, provided both sides are defined:

$(x+1)^2 = 2x^2 + 8x$

Step 3: Expand and simplify the equation:

$(x+1)^2 = x^2 + 2x + 1$

So, now the equation becomes:

$x^2 + 2x + 1 = 2x^2 + 8x$

Rearranging gives:

$x^2 + 2x + 1 - 2x^2 - 8x = 0$

Which simplifies to:

$-x^2 - 6x + 1 = 0$

Or multiplying through by -1:

$x^2 + 6x - 1 = 0$

Step 4: Solve the quadratic equation using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, with $a = 1$, $b = 6$, and $c = -1$.

$x = \frac{-6 \pm \sqrt{36 + 4}}{2} = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2} = -3 \pm \sqrt{10}$

Step 5: Verify possible solutions by checking the domain. For $x = -3 + \sqrt{10}$, both $x+1 > 0$ and $2x^2 + 8x > 0$ are satisfied. For $x = -3 - \sqrt{10}$, $x+1$ would be negative, violating the logarithm domain.

Therefore, the solution to the problem is $x = -3 + \sqrt{10}$.