To solve the equation $\log_6 x + \log_6 (x + 3) = 9$, we can use the properties of logarithms. Here's the step-by-step solution:

Step 1: Combine the logarithmic terms

Using the property $\log_a m + \log_a n = \log_a (m \cdot n)$, we combine the terms: $\log_6 (x \cdot (x + 3)) = 9$

This simplifies to: $\log_6 (x^2 + 3x) = 9$

Step 2: Convert the logarithmic equation to exponential form

The equation $\log_a b = c$ is equivalent to $a^c = b$. Applying this: $6^9 = x^2 + 3x$

Step 3: Simplify the exponential term

Calculating $6^9$: $6^9 = 10077696$ Thus, the equation becomes: $x^2 + 3x - 10077696 = 0$

Step 4: Solve the quadratic equation

The quadratic equation is: $x^2 + 3x - 10077696 = 0$

We solve this using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 3$, and $c = -10077696$. Substituting: $x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10077696)}}{2(1)}$ $x = \frac{-3 \pm \sqrt{9 + 40310784}}{2}$ $x = \frac{-3 \pm \sqrt{40310793}}{2}$

Step 5: Approximate the roots

Calculating $\sqrt{40310793}$ gives approximately $6348.28$. Thus: $x = \frac{-3 + 6348.28}{2} \quad \text{or} \quad x = \frac{-3 - 6348.28}{2}$ $x = \frac{6345.28}{2} \quad \text{or} \quad x = \frac{-6351.28}{2}$ $x = 3172.64 \quad \text{or} \quad x = -3175.64$

Step 6: Discard the negative solution

Since $x > 0$ (logarithms are undefined for non-positive values), the solution is: $x \approx 3172.64$

Final Answer:

$x \approx 3172.64$

Would you like further clarification or additional steps explained?

Related Questions:

1. How does the quadratic formula work in general for solving equations?
2. Why must $x > 0$ for logarithmic functions?
3. How is the property $\log_a m + \log_a n = \log_a (m \cdot n)$ derived?
4. What is the significance of converting a logarithmic equation to exponential form?
5. How can $6^9$ be computed efficiently?

Tip:

Always check the domain of the logarithmic function before solving; ensure that the arguments are positive!