To solve the logarithmic equation:

$2\log(3x + 2) = \log 12$

Step 1: Apply logarithmic properties

Use the property of logarithms that states $a\log(b) = \log(b^a)$. Applying this property to the left-hand side:

$2\log(3x + 2) = \log((3x + 2)^2)$

This gives us the equation:

$\log((3x + 2)^2) = \log 12$

Step 2: Eliminate the logarithms

Since the logarithms have the same base (base 10), we can equate the arguments:

$(3x + 2)^2 = 12$

Step 3: Solve the quadratic equation

Now, solve for $x$. First, take the square root of both sides:

$3x + 2 = \pm \sqrt{12}$

Simplify the square root:

$\sqrt{12} = 2\sqrt{3}$

So, the equation becomes:

$3x + 2 = \pm 2\sqrt{3}$

Step 4: Solve for $x$

Now, solve for $x$ in both cases:

1. For $3x + 2 = 2\sqrt{3}$:

$3x = 2\sqrt{3} - 2$

$x = \frac{2\sqrt{3} - 2}{3}$

2. For $3x + 2 = -2\sqrt{3}$:

$3x = -2\sqrt{3} - 2$

$x = \frac{-2\sqrt{3} - 2}{3}$

Step 5: Check for extraneous solutions

Logarithms are only defined for positive arguments. This means that $3x + 2 > 0$, or $x > -\frac{2}{3}$. So, we need to check if either solution satisfies this condition. After evaluating:

• $x = \frac{2\sqrt{3} - 2}{3}$ is valid.
• $x = \frac{-2\sqrt{3} - 2}{3}$ is not valid because it results in a negative value for $3x + 2$.

Final Solution:

$x = \frac{2\sqrt{3} - 2}{3}$

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Would you like more details or have any questions about this solution?

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Here are 5 questions to explore further:

1. How do you apply properties of logarithms in equations?
2. What are the conditions for logarithmic functions to be valid?
3. How do you simplify expressions involving square roots?
4. Can you have negative arguments inside a logarithm?
5. What is an extraneous solution, and why do they occur in logarithmic equations?

Tip: Always check for extraneous solutions when solving logarithmic equations by ensuring the arguments of the logarithms remain positive.