Let's solve the given problem step-by-step.

Step 1: Simplify using logarithmic properties

Given equation:

$2\log x = \log 2 + \log (3x - 4)$

Using the properties of logarithms:

1. The property $a \log b = \log b^a$ transforms $2 \log x$ into:

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

2. The property $\log a + \log b = \log (a \cdot b)$ transforms $\log 2 + \log (3x - 4)$ into:

   $\log 2 + \log (3x - 4) = \log (2 \cdot (3x - 4))$

So, the equation simplifies to:

$\log (x^2) = \log (2(3x - 4))$

Step 2: Set the arguments equal

Since the logarithms are equal, we can set the arguments equal:

$x^2 = 2(3x - 4)$

Step 3: Expand and simplify

Expand the right side:

$x^2 = 6x - 8$

Rearrange terms to form a quadratic equation:

$x^2 - 6x + 8 = 0$

Step 4: Solve the quadratic equation

To solve $x^2 - 6x + 8 = 0$, use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -6$, and $c = 8$.

Substituting these values:

$x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}$

$x = \frac{6 \pm \sqrt{36 - 32}}{2}$

$x = \frac{6 \pm \sqrt{4}}{2}$

$x = \frac{6 \pm 2}{2}$

This gives two solutions:

$x_1 = \frac{6 + 2}{2} = 4, \quad x_2 = \frac{6 - 2}{2} = 2$

Step 5: Check the validity of solutions

Since the arguments inside the logarithms must be positive:

1. For $x = 4$:
   • $\log(3x - 4) = \log(3(4) - 4) = \log(8)$, which is valid.
2. For $x = 2$:
   • $\log(3x - 4) = \log(3(2) - 4) = \log(2)$, which is also valid.

Thus, both $x = 4$ and $x = 2$ are valid solutions.

Final Answer

$x = 4 \text{ and } x = 2$

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Would you like more details on any of the steps or have any questions? Here are some related questions you might find interesting:

1. How do you verify solutions of a logarithmic equation?
2. What are the properties of logarithms used in solving equations?
3. How do you solve quadratic equations using factoring?
4. Can logarithmic equations have extraneous solutions?
5. What is the domain of logarithmic functions?

Tip: Always check the arguments inside logarithms for validity when solving logarithmic equations to avoid extraneous solutions!