We are given the equation:

$2\log(3x + 2) = \log(121)$

Step 1: Simplify the equation

Using the property of logarithms $a\log(b) = \log(b^a)$, we rewrite the left-hand side:

$\log((3x + 2)^2) = \log(121)$

Step 2: Remove the logarithms

Since both sides of the equation have logarithms, we can set the arguments equal to each other:

$(3x + 2)^2 = 121$

Step 3: Solve for $x$

Now, take the square root of both sides:

$3x + 2 = \pm \sqrt{121}$ $3x + 2 = \pm 11$

This gives us two possible equations:

1. $3x + 2 = 11$
2. $3x + 2 = -11$

Step 4: Solve both equations

1. $3x + 2 = 11$ $3x = 11 - 2 = 9$ $x = \frac{9}{3} = 3$

2. $3x + 2 = -11$ $3x = -11 - 2 = -13$ $x = \frac{-13}{3} = -\frac{13}{3}$

Step 5: Verify the solutions

We need to check if these values of $x$ make sense in the original logarithmic equation.

• For $x = 3$: $3x + 2 = 3(3) + 2 = 11 > 0$ So, $x = 3$ is valid.

• For $x = -\frac{13}{3}$: $3x + 2 = 3\left(-\frac{13}{3}\right) + 2 = -13 + 2 = -11 < 0$ Since the argument of the logarithm cannot be negative, $x = -\frac{13}{3}$ is not valid.

Final Answer:

The valid solution is: $x = 3$

Would you like further details, or have any questions?

Relative questions:

1. How do logarithmic properties help simplify equations?
2. Can you explain why negative values inside a logarithm are not allowed?
3. How do we deal with the square root in an equation like this?
4. What happens if we have different bases in the logarithms?
5. What other methods can be used to solve logarithmic equations?

Tip:

Always check the domain of the logarithmic function before finalizing solutions.