We are tasked with solving the equation:

$\ln(x + 1) + \ln(x) = \ln(2)$

Step 1: Use the logarithmic property

We can use the logarithmic property:

$\ln(a) + \ln(b) = \ln(a \cdot b)$

Thus, the left-hand side of the equation becomes:

$\ln((x + 1) \cdot x) = \ln(2)$

This simplifies to:

$\ln(x(x + 1)) = \ln(2)$

Step 2: Remove the natural logarithms

Since the natural logarithm functions on both sides of the equation are equal, the arguments must be equal. Therefore:

$x(x + 1) = 2$

Step 3: Expand the equation

Now, expand the left-hand side:

$x^2 + x = 2$

Step 4: Rearrange the equation into a quadratic form

Move all terms to one side to form a standard quadratic equation:

$x^2 + x - 2 = 0$

Step 5: Solve the quadratic equation

We can solve this quadratic equation using factoring. We look for two numbers that multiply to $-2$ and add to $1$. These numbers are $2$ and $-1$, so we factor the quadratic as:

$(x - 1)(x + 2) = 0$

Step 6: Find the solutions

Set each factor equal to zero:

$x - 1 = 0 \quad \text{or} \quad x + 2 = 0$

This gives us the solutions:

$x = 1 \quad \text{or} \quad x = -2$

Step 7: Check for extraneous solutions

Since the logarithmic function is only defined for positive arguments, we must check the domain of the original equation. Both $\ln(x + 1)$ and $\ln(x)$ require $x > 0$. Therefore, the solution $x = -2$ is not valid.

Thus, the only valid solution is:

$x = 1$

Would you like further clarification on any step?

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Here are 5 related questions to deepen your understanding:

1. How do we solve logarithmic equations with different bases?
2. What is the domain of a logarithmic function?
3. Can logarithmic equations have no solutions? If so, under what conditions?
4. How can we solve a quadratic equation that cannot be factored easily?
5. What are the key properties of logarithms that help in solving equations?

Tip: Always check the domain of logarithmic functions to ensure the solution is valid!