- Establish the domain for \(\log_2(x) + \log_2(x+7) = 3\) by identifying that \(x > 0\).
- Solve the equation to find candidates \(x = -8\) and \(x = 1\).
- Exclude \(x = -8\) as it is outside the domain, leaving only \(x = 1\).
- The correct option is \(\boxed{\text{a}}\).
Explanation
- Domain Analysis
We are given the logarithmic equation \(\log_2(x) + \log_2(x+7) = 3\). To solve this correctly, we must first establish the domain of the equation. Logarithmic functions \(\log_b(y)\) are only defined for arguments where \(y > 0\). In our equation, this requires that \(x > 0\) and \(x + 7 > 0\). Combining these, the domain for the equation is \(x > 0\).
- Solving the Algebraic Equation
Using the properties of logarithms, we can combine the terms: \(\log_2(x) + \log_2(x+7) = 3\) becomes \(\log_2(x(x+7)) = 3\). Converting the logarithmic equation to an exponential form gives \(x(x+7) = 2^3\), which simplifies to \(x^2 + 7x = 8\), or \(x^2 + 7x - 8 = 0\). Factoring the quadratic expression, we get \((x+8)(x-1) = 0\), which yields the potential candidates \(x = -8\) and \(x = 1\).
- Verifying Solutions and Conclusion
Now, we must test the candidate solutions \(x = -8\) and \(x = 1\) against our domain condition \(x > 0\). For \(x = -8\), the condition is not satisfied because \(-8\) is not greater than 0. For \(x = 1\), the condition is satisfied since \(1 > 0\). Substituting \(x = 1\) back into the original equation, we get \(\log_2(1) + \log_2(1+7) = 0 + \log_2(8) = 0 + 3 = 3\), confirming it is indeed a valid solution. Therefore, the only true solution is \(x = 1\). Comparing this result with the provided options, we conclude that option (a) is correct.
Examples
Logarithmic equations are used in various fields like acoustics and chemistry. For instance, the pH level of a substance is calculated using a logarithmic scale, and the intensity of sound in decibels is measured using logs. When solving these problems in science, verifying your results within the domain is essential, just as we did here, because you cannot take the logarithm of a negative number in real-world measurements. Understanding these restrictions ensures your calculations remain physically meaningful.