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When encountering equations that involve logarithms, the logarithm product rule is a handy tool for simplifying expressions. This rule states that the sum of two logarithms with the same base can be combined into a single log expression by converting the sum into a product. In other words, if you have \(\text{{log}}_a(m) + \text{{log}}_a(n)\), you can combine them into one logarithm through multiplication of the arguments: \(\text{{log}}_a(m \times n)\).

Let's apply this rule to an exercise where we need to solve for \(x\) in the equation \(\text{{log}}_2(x-1) + \text{{log}}_2(x+1) = 3\). By leveraging the logarithm product rule, the equation simplifies to \(\text{{log}}_2((x-1)(x+1)) = 3\), which is the first crucial step in finding the value of \(x\). Without understanding this rule, solving such logarithmic equations would become much more challenging.

Exponentiation is the process of raising a number to a certain power, which serves as the inverse operation of logarithms. In simpler terms, if you have an equation with a logarithm such as \(\text{{log}}_a(b)\), and you want to 'cancel out' the logarithm, you can raise the base of the logarithm, \(a\), to the power of the entire log expression which will leave you with just \(b\). This is represented by \(a^{\text{{log}}_a(b)} = b\).

In our exercise, once we've combined the logarithms using the product rule, we are faced with the equation \(\text{{log}}_2((x-1)(x+1)) = 3\). To remove the logarithm and solve for \(x\), we exponentiate both sides by the base of the logarithm, which gives us \(2^{\text{{log}}_2((x-1)(x+1))} = 2^3\). This simplifies to \((x-1)(x+1) = 8\), effectively using exponentiation to move forward in finding the solution to the equation.

Quadratic equations are polynomial equations of the second degree, which generally take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Solving these equations often requires factoring, completing the square, or applying the quadratic formula. A key characteristic of quadratic equations is that they can have either one or two real solutions, also known as roots.

In the context of our exercise, when we remove the logarithm through exponentiation, the equation transforms into a quadratic form \(x^2 - 1 = 8\). To find the value of \(x\), we start by adding 1 to each side of the equation to isolate the quadratic term, resulting in \(x^2 = 9\). Finally, we find the square root of both sides, which provides two potential solutions: \(x = \pm 3\). However, due to the nature of logarithms (they are not defined for negative arguments), we eliminate \(x = -3\) as a viable solution, leaving us with the only solution: \(x = 3\). Understanding how to manipulate and solve quadratic equations is essential in algebra and is the final step in solving the given exercise.