To solve the equation:

$\log_{10}(x10(x + 5)) = 1 + \log_{10}(2),$

follow these steps:

Step 1: Simplify the left-hand side

Using the property of logarithms $\log_{10}(ab) = \log_{10}(a) + \log_{10}(b)$, rewrite the left-hand side: $\log_{10}(x10(x + 5)) = \log_{10}(x) + \log_{10}(10) + \log_{10}(x + 5).$

Since $\log_{10}(10) = 1$, this becomes: $\log_{10}(x10(x + 5)) = \log_{10}(x) + 1 + \log_{10}(x + 5).$

Step 2: Set up the equation

Substituting back into the original equation: $\log_{10}(x) + 1 + \log_{10}(x + 5) = 1 + \log_{10}(2).$

Cancel $1$ from both sides: $\log_{10}(x) + \log_{10}(x + 5) = \log_{10}(2).$

Step 3: Combine logarithmic terms

Using the property $\log_{10}(a) + \log_{10}(b) = \log_{10}(ab)$, combine the terms: $\log_{10}(x(x + 5)) = \log_{10}(2).$

Step 4: Exponentiate to remove the logarithm

Exponentiating both sides with base 10: $x(x + 5) = 2.$

Step 5: Solve the quadratic equation

Expand and simplify: $x^2 + 5x - 2 = 0.$

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = 5$, $c = -2$. Substituting: $x = \frac{-5 \pm \sqrt{5^2 - 4(1)(-2)}}{2(1)} = \frac{-5 \pm \sqrt{25 + 8}}{2} = \frac{-5 \pm \sqrt{33}}{2}.$

Step 6: Consider only valid solutions

Since $x > 0$ (logarithms are undefined for non-positive values), the solution is: $x = \frac{-5 + \sqrt{33}}{2}.$

Final Answer:

$x = \frac{-5 + \sqrt{33}}{2}.$

Would you like a detailed breakdown of any step or further assistance?

Questions:

1. How does the quadratic formula work in other contexts?
2. Why can logarithms only accept positive arguments?
3. How does combining logarithms simplify expressions?
4. What happens if no real solutions exist in a logarithmic equation?
5. How can we interpret this solution geometrically?

Tip:

When solving logarithmic equations, always check if the solution satisfies the domain of the logarithm (positive arguments only).