We are given the following logarithmic equation:

$\log_{\sqrt{6}}(3x) + \log_{\sqrt{6}}(x+1) = 2$

Step 1: Use the logarithmic property for addition

The property of logarithms states that:

$\log_b(A) + \log_b(B) = \log_b(AB)$

Thus, we can combine the two logarithms:

$\log_{\sqrt{6}}(3x(x+1)) = 2$

Step 2: Convert the logarithmic equation to an exponential form

The equation $\log_b(A) = C$ can be rewritten as $A = b^C$. Applying this to our equation, we get:

$3x(x+1) = (\sqrt{6})^2$

Since $(\sqrt{6})^2 = 6$, the equation becomes:

$3x(x+1) = 6$

Step 3: Solve the resulting quadratic equation

Expand the left-hand side:

$3x(x + 1) = 3x^2 + 3x$

Now, set the equation equal to 6:

$3x^2 + 3x = 6$

Simplify:

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

Divide through by 3:

$x^2 + x - 2 = 0$

Factor the quadratic equation:

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

Step 4: Solve for $x$

Set each factor equal to zero:

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

Solving these gives:

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

Step 5: Check for extraneous solutions

Since logarithms are only defined for positive arguments, we need to check both possible values of $x$.

• For $x = -2$, the terms $\log_{\sqrt{6}}(3(-2))$ and $\log_{\sqrt{6}}((-2)+1)$ involve logarithms of negative numbers, which are undefined. Therefore, $x = -2$ is not a valid solution.

• For $x = 1$, both arguments inside the logarithms are positive ($3x = 3$ and $x+1 = 2$), so this is a valid solution.

Final Answer:

The solution to the equation is:

$x = 1$

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Would you like more details on any of the steps, or have any other questions? Here are some related questions:

1. What are the properties of logarithms used to solve this equation?
2. How do you handle equations involving different bases in logarithms?
3. How can we check for extraneous solutions in logarithmic equations?
4. What is the general method for solving logarithmic equations?
5. How do you simplify logarithmic expressions when the base is a square root?

Tip: Always check for extraneous solutions when solving logarithmic and exponential equations!