91Ó°ÊÓ

Logarithmic expressions are the inverses of exponential functions. When we encounter an equation with logarithms, such as the one given in this exercise, understanding the properties of logarithms is crucial. The key property utilized in this exercise is the logarithm product rule, which states that the sum of two logarithms with the same base, like \( \log_b(m) + \log_b(n) \), can be combined into a single logarithm by multiplying the arguments, leading to \( \log_b(mn) \).

When solving logarithmic equations, it's important to remember that the argument of the logarithm—what's inside the logarithm—must always be greater than zero. This is because logarithms are undefined for non-positive numbers. In the original equation \( \log_{6}(x+3) + \log_{6}(x+4) = 1 \), we are implicitly assuming that \( x+3 > 0 \) and \( x+4 > 0 \) to ensure the expressions are within the domain of real numbers. Using these rules and understanding the domain requirements are the first steps to solving any logarithmic equation.

Exponential functions are fundamental in solving logarithmic equations, as they are the inverse operations. In the example provided, after combining the logarithmic expressions, we convert the logarithmic equation into an exponential equation by remembering that \( \log_b(y) = x \) is equivalent to \( b^x = y \). This allows us to transform the logarithmic form \( \log_{6}((x+3)(x+4)) = 1 \) into the exponential form \( 6^1 = (x+3)(x+4) \).

This step is crucial because it converts the problem into a quadratic equation, which often makes it easier to solve. By understanding how to move between logarithmic and exponential forms, one is better equipped to tackle various logarithmic equations and apply the correct algebraic techniques to find the solution.

When faced with a quadratic equation such as \( x^2 + 7x + 6 = 0 \) in the context of logarithmic equations, the quadratic formula is an invaluable tool. It provides a consistent method for finding the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In the provided exercise, the quadratic equation is \( x^2 + 7x + 6 = 0 \) which can be solved by plugging the coefficients \( a = 1 \) , \( b = 7 \) , and \( c = 6 \) into the formula. Upon solving, it produces two potential solutions. However, not all solutions derived from the quadratic formula will satisfy the original logarithmic equation. Solutions that make the argument of any logarithm negative or zero are extraneous and must be rejected. In this case, \( x = -6 \) is rejected, leaving \( x = -1 \) as the valid solution.