We need to solve the equation:

$\log(x) + \log(x + 2) = 4.$

Step-by-Step Solution:

1. Combine the Logarithmic Terms: Using the logarithmic property $\log(a) + \log(b) = \log(a \cdot b)$, we combine the two terms: $\log(x) + \log(x + 2) = \log(x(x + 2)).$ This simplifies the equation to: $\log(x(x + 2)) = 4.$

2. Rewrite in Exponential Form: Using the property $\log_a(b) = c \implies b = a^c$, we rewrite: $x(x + 2) = 10^4.$ Since $10^4 = 10000$, we now have: $x(x + 2) = 10000.$

3. Expand and Rearrange: Expand the left-hand side: $x^2 + 2x = 10000.$ Rearrange into standard quadratic form: $x^2 + 2x - 10000 = 0.$

4. Solve the Quadratic Equation: Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = 2$, and $c = -10000$. Substituting these values: $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-10000)}}{2(1)}.$ Simplify: $x = \frac{-2 \pm \sqrt{4 + 40000}}{2}.$ $x = \frac{-2 \pm \sqrt{40004}}{2}.$

5. Approximate the Square Root: Calculate $\sqrt{40004}$: $\sqrt{40004} \approx 200.01.$ Substituting this: $x = \frac{-2 \pm 200.01}{2}.$

6. Calculate Both Solutions: $x = \frac{-2 + 200.01}{2} \quad \text{or} \quad x = \frac{-2 - 200.01}{2}.$ $x = \frac{198.01}{2} = 99.005 \quad \text{or} \quad x = \frac{-202.01}{2} = -101.005.$

7. Exclude the Negative Solution: Since $\log(x)$ is undefined for $x \leq 0$, we discard $x = -101.005$.

Final Answer:

$x \approx 99.005.$

Would you like further clarification or additional steps explained?

Related Questions:

1. How do we handle equations with more than two logarithmic terms?
2. What happens if $x + 2$ equals zero in this scenario?
3. How would changing the base of the logarithm affect the solution?
4. Can logarithmic equations have multiple valid solutions?
5. How does graphing $\log(x) + \log(x + 2)$ help visualize the solution?

Tip:

Always verify the solution by substituting it back into the original logarithmic equation to ensure it satisfies all conditions.