We are tasked with solving the logarithmic equation:

$\log_4{(x+6)} + \log_4{(x)} = 2$

Step 1: Use the logarithmic product rule

The logarithmic product rule states that:

$\log_b{(a)} + \log_b{(c)} = \log_b{(a \cdot c)}$

Applying this rule to our equation:

$\log_4{[(x+6) \cdot x]} = 2$

Simplifying the expression inside the logarithm:

$\log_4{(x^2 + 6x)} = 2$

Step 2: Rewrite the logarithmic equation in exponential form

The logarithmic equation $\log_b{(a)} = c$ is equivalent to $b^c = a$. Applying this property:

$4^2 = x^2 + 6x$

This simplifies to:

$16 = x^2 + 6x$

Step 3: Rearrange into a standard quadratic equation

Move all terms to one side of the equation:

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

Step 4: Solve the quadratic equation

We now solve $x^2 + 6x - 16 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $a = 1$, $b = 6$, and $c = -16$, we have:

$x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-16)}}{2(1)}$ $x = \frac{-6 \pm \sqrt{36 + 64}}{2}$ $x = \frac{-6 \pm \sqrt{100}}{2}$ $x = \frac{-6 \pm 10}{2}$

Thus, the two possible solutions are:

$x = \frac{-6 + 10}{2} = \frac{4}{2} = 2$ $x = \frac{-6 - 10}{2} = \frac{-16}{2} = -8$

Step 5: Check for extraneous solutions

Logarithms are only defined for positive arguments, so we need to check the solutions:

• For $x = 2$, both $\log_4{(2+6)}$ and $\log_4{(2)}$ are valid.
• For $x = -8$, $\log_4{(-8)}$ is not defined, so this solution is extraneous.

Final Solution:

The only valid solution is:

$x = 2$

Would you like further details, or do you have any questions?

Related Questions:

1. How do you graph logarithmic functions?
2. What are the other logarithmic rules besides the product rule?
3. Can we solve logarithmic equations with different bases?
4. How do you simplify logarithmic expressions involving radicals?
5. What are the conditions for extraneous solutions in logarithmic equations?

Tip:

When solving logarithmic equations, always check if the arguments of the logarithms are positive, as logarithms are not defined for negative numbers or zero.