91Ó°ÊÓ

Understanding the logarithmic product rule is crucial when dealing with equations involving multiple logarithmic terms. Simplifying the equation by combining logarithms not only makes the problem neater but also easier to solve. In our exercise, the rule is applied to condense two log terms into one. The rule states that the sum of two logarithms with the same base equals the logarithm of the product of their arguments.

Therefore, \( \log_{b}(m) + \log_{b}(n) = \log_{b}(mn) \). This property lets us combine \( \log_{5}(x+4) \) and \( \log_{5}(x+2) \) into \( \log_{5}((x+4)(x+2)) \), setting up the stage for the next step in solving the equation.

Once the logarithms are combined, the equation takes on a simpler form that can be converted into exponential form. This process helps in extracting the variable from the logarithm. The equation \( \log_{b}(x) = y \) implies that \( b^y = x \), where \( b \) is the base of the logarithm, \( y \) is the logarithmic value, and \( x \) is the argument. Using this logic, we converted \( \log_{5}((x+4)(x+2)) = 2 \) into \( 5^2 = (x+4)(x+2) \), allowing us to proceed with solving for \( x \) algebraically in the following steps.

With the equation now in quadratic form \( x^2 + 6x - 17 = 0 \), we turn to the quadratic formula to find the roots of the equation. The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), gives us the solutions to any quadratic equation \( ax^2 + bx + c = 0 \). In our problem, the values of \( a \), \( b \), and \( c \) are 1, 6, and -17, respectively. After substituting these values into the formula, we obtain two potential solutions for \( x \). It's important for students to remember that while this method always works, it is worth looking for simpler factoring options before using this formula. Nonetheless, it provides a surefire way to find the solutions to quadratic equations that might not easily be factored.