Verified Answer :
- The domain requirement is \(x > 3\).
- Using the logarithmic product property gives \(\log_2[(x+1)(x-3)] = 3\).
- Converting the logarithmic equation to exponential form gives \(2^3 = (x+1)(x-3)\).
- The true statements are that the equation converts to \(2^3=(x+1)(x-3)\) and the product property gives \(\log_2[(x+1)(x-3)]=3\). The solution \(x=1+2\sqrt{3}\) makes the other statements false.
Explanation
- Problem Analysis and Domain Constraints
To solve the equation \(\log_2(x+1) + \log_2(x-3) = 3\), we first identify the domain constraints. For the logarithmic functions to be defined, the arguments must be strictly positive: \(x+1 > 0\) and \(x-3 > 0\). This implies \(x > -1\) and \(x > 3\), so we must have \(x > 3\).
- Applying Logarithmic Properties
Next, we apply the product property of logarithms, which states that \(\log_b(M) + \log_b(N) = \log_b(MN)\). Applying this to our equation gives \(\log_2[(x+1)(x-3)] = 3\). This confirms the statement that the product property gives \(\log_2[(x+1)(x-3)] = 3\).
- Converting to an Algebraic Equation
Using the definition of a logarithm, where \(\log_b(Y) = Z\) is equivalent to \(b^Z = Y\), we can convert the equation \(\log_2[(x+1)(x-3)] = 3\) into \(2^3 = (x+1)(x-3)\). This confirms the statement that this converts to \(2^3 = (x+1)(x-3)\). Evaluating the right side, we get \((x+1)(x-3) = x^2 - 3x + x - 3 = x^2 - 2x - 3\). Thus, the equation becomes \(8 = x^2 - 2x - 3\), which simplifies to \(x^2 - 2x - 11 = 0\). Comparing this to the statement \(x^2 - 2x - 3 = 16\), we see that the latter is incorrect.
- Solving and Verifying Solutions
Solving \(x^2 - 2x - 11 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we get \(x = \frac{2 \pm \sqrt{4 - 4(1)(-11)}}{2} = \frac{2 \pm \sqrt{48}}{2} = 1 \pm 2\sqrt{3}\). Since the domain requires \(x > 3\), we discard the negative root \(1 - 2\sqrt{3}\). The only solution is \(x = 1 + 2\sqrt{3} \approx 4.46\). This contradicts the statement that \(x=5\) is the solution, and that both \(x=5\) and \(x=-3\) are valid. Therefore, the only true statements are the first and fourth ones.
Examples
Logarithmic equations are used in various fields such as chemistry to determine pH levels, in finance to calculate interest rates, and in acoustics to measure sound intensity in decibels. For example, if the noise level in a specific environment grows following a logarithmic model, solving such equations helps engineers determine when noise levels might exceed safety thresholds, enabling the design of appropriate mitigation strategies like soundproofing.