91Ó°ÊÓ

When solving equations involving logarithms, quadratic equations often surface as an intermediary step. A quadratic equation is a second-order polynomial equation of the form \(ax^2 + bx + c = 0\). In this scenario, after combining the logarithmic terms, we expand and rearrange the equation into a quadratic form: \(x^2 - 3x - 10 = 0\). This equation can be solved using several methods:

• Factoring: Decompose the quadratic expression into the product of two binomials, \((x - 5)(x + 2) = 0\), which is the approach taken in this solution.
• Completing the Square: Transform the equation into a perfect square trinomial.
• Quadratic Formula: Use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots directly.

Choosing a method depends on the complexity of the equation and personal preference, but all methods lead to finding the values of \(x\) that satisfy the equation. In our case, the factored form gives solutions \(x = 5\) and \(x = -2\).

To solve logarithmic equations, converting them into exponential form is a helpful step. Logarithms and exponents are inverse operations. If you have \(\log_b(a) = c\), it translates to \(b^c = a\) in exponential form. In our problem, the logarithmic equation \(\log(x(x-3)) = 1\) implies that \(10^1 = x(x-3)\) because the logarithm is base 10 by default if not specified.

This transformation allows us to remove the logarithmic function and deal with a polynomial equation instead. Once in exponential form, the equation becomes straightforward to manipulate and solve, leading us into quadratic territory as seen in the previous solution step. Utilizing exponential conversion skills is vital for efficiently handling and simplifying complex logarithmic equations.

Logarithmic properties are indispensable tools when working with equations that include logarithmic terms. In the given problem, we utilized the logarithmic product rule that states \(\log_b(M) + \log_b(N) = \log_b(M \times N)\). This property allowed us to combine \(\log x + \log (x-3)\) into one single log term: \(\log(x(x-3))\).

Understanding the basic log laws like the product, quotient, and power rules can dramatically simplify and solve logarithmic expressions and equations. By transforming multiple log terms into combined forms, we can then convert them into a more workable format using exponential expressions. Paying attention to these properties helps with recognizing valid manipulations required to progress with problem-solving in logarithmic contexts.

Domain restrictions are crucial when dealing with logarithmic equations. Logs are only defined for positive arguments, which means every expression inside a logarithm must remain positive for real number solutions. In our exercise, this requirement immediately eliminates solutions like \(x = -2\), which would cause the arguments \(x\) and \(x-3\) to be negative in \(\log x\) and \(\log (x-3)\), respectively.

Understanding the domain of logarithmic functions helps accurately determine valid solutions. For instance, for \(\log x + \log (x-3)\) to be defined, both \(x > 0\) and \(x - 3 > 0\) must be true. Therefore, the values of \(x\) must be greater than 3. Only the solution \(x = 5\) satisfies all domain restrictions, making it the correct and valid solution for this logarithmic equation.

Always remember to verify solutions against any domain conditions to ensure they are mathematically valid and meaningful.