To solve the equation:

$\log_4(x^2 - 6x) = 2$

Step 1: Rewrite the Logarithmic Equation in Exponential Form

The equation can be rewritten in exponential form as:

$x^2 - 6x = 4^2$

since $4^2 = 16$, this becomes:

$x^2 - 6x = 16$

Step 2: Rearrange into a Standard Quadratic Equation

Move all terms to one side:

$x^2 - 6x - 16 = 0$

Step 3: Solve the Quadratic Equation

Now we factor the quadratic:

$(x - 8)(x + 2) = 0$

Setting each factor equal to zero:

$x - 8 = 0 \Rightarrow x = 8$ $x + 2 = 0 \Rightarrow x = -2$

Step 4: Check for Valid Solutions

Since we are dealing with a logarithmic function, the argument $x^2 - 6x$ must be positive.

1. For $x = 8$:
   $x^2 - 6x = 8^2 - 6 \cdot 8 = 64 - 48 = 16$, which is positive.

2. For $x = -2$:
   $x^2 - 6x = (-2)^2 - 6 \cdot (-2) = 4 + 12 = 16$, which is also positive.

Solution

Both values, $x = 8$ and $x = -2$, satisfy the original equation.

$\boxed{x = 8 \text{ and } x = -2}$

Would you like further explanation on logarithmic functions, or do you have any questions?

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Related Questions:

1. How do you solve logarithmic equations with different bases?
2. What is the importance of checking domain constraints in logarithmic equations?
3. How do you factor a quadratic equation if it doesn't factor nicely?
4. What if the logarithmic function has a base different than an integer?
5. How do you solve exponential equations?

Tip: Always check the argument of a logarithmic function to ensure it stays positive for real solutions.