Which of the following shows the extraneous solution to the logarithmic equation below?
$2 \log _5(x+1)=2$
To solve the logarithmic equation \(2 \log_5(x+1) = 2\), we first define its domain. The logarithmic function \(\log_5(u)\) is defined only for \(u > 0\). Thus, the argument \(x+1\) must satisfy the condition \(x+1 > 0\), which implies \(x > -1\). Any value of \(x\) that is less than or equal to \(-1\) will result in a non-positive argument, making the logarithmic expression undefined in the real number system.
We simplify the given equation \(2 \log_5(x+1) = 2\) by dividing both sides by 2, which yields \(\log_5(x+1) = 1\). Converting this logarithmic equation to its exponential form, we have \(x+1 = 5^1\), which simplifies to \(x+1 = 5\). Solving for \(x\), we find the only valid solution is \(x = 4\). Since \(4 > -1\), this solution is within the domain of the original equation.
An extraneous solution is a value that emerges as a potential solution from algebraic manipulation but does not satisfy the original equation's domain constraints. If we were to manipulate the equation \(\log_5(x+1) = 1\) into an equivalent form such as \(\log_5((x+1)^2) = 2\), squaring the argument \((x+1)^2 = 25\) yields two possible values: \(x+1 = 5\) (giving \(x=4\)) and \(x+1 = -5\) (giving \(x=-6\)). While \(x = -6\) satisfies the squared form of the equation because \((-6+1)^2 = 25\), it makes the original expression \(\log_5(x+1)\) undefined. Therefore, \(x = -6\) is an extraneous solution.
Extraneous solutions are common when solving equations involving square roots or logarithms. For example, if an engineer is calculating the dimensions of a structural component and solves an equation involving a square root, they might find a negative value that works algebraically but is physically impossible (like a negative length). Checking domain constraints ensures the solution is both mathematically valid and physically meaningful, preventing errors in real-world engineering or construction projects.