91Ó°ÊÓ

A quadratic equation is a second-degree polynomial equation in the form of \(ax^2 + bx + c = 0\). In the provided exercise, after setting the arguments of the logarithms equal to each other, we get the equation \(2x = 24 - x^2\). Rearranging this gives us a quadratic form: \(x^2 + 2x - 24 = 0\).
The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) is used to find the roots (or solutions) of the quadratic equation. Here, the coefficients are \(a = 1\), \(b = 2\), and \(c = -24\). Substituting these values into the quadratic formula, we solve for \(x\) and obtain two solutions: \(x = 4\) and \(x = -6\).

Remember, solving a quadratic equation means finding the values of \(x\) that make the equation true. These values are called the roots of the equation.

Logarithms have several properties that can be very useful in solving equations. One important property is that if the logarithms on both sides of the equation have the same base, the arguments can be set equal to each other. This is known as the 'one-to-one property' of logarithms.

In our exercise, since \( \log_{3}(2x) = \log_{3}(24 - x^2) \), we set the arguments equal: \( 2x = 24 - x^2 \). This property helps to simplify the logarithmic equation into a polynomial equation, which can then be solved using standard algebraic methods.

Furthermore, logarithms have the base-argument relationship that defines them: \( \log_b(a) = c \) means that \(b^c = a\). Understanding these foundational properties makes working with logarithms more straightforward and helps in solving complex-logarithmic equations efficiently.

When solving equations involving logarithms or other functions, sometimes solutions arise that do not satisfy the original equation. These are known as extraneous solutions. They occur due to the algebraic manipulations used in the solving process, especially when dealing with squared terms or logarithms.

In our exercise, we found two solutions: \(x = 4\) and \(x = -6\). To verify which solutions are valid, we substitute them back into the original equation. For \(x = 4\): \( \log_{3}(2 \times 4) = \log_{3}(24 - 4^2) \) which simplifies to \( \log_{3}(8) = \log_{3}(8) \), confirming this as a valid solution.

For \(x = -6\), substituting it back gives \( \log_{3}(2 \times (-6)) \), leading to a negative argument \(\log_{3}(-12)\), which is not valid because the logarithm of a negative number is undefined in real numbers. Thus, \(x = -6\) is an extraneous solution.
Always check for extraneous solutions to ensure your final answers are correct according to the original equation.