91Ó°ÊÓ

The logarithm product rule is a handy tool when working with expressions involving logarithms. It states that the sum of two logarithms is equal to the logarithm of their product. Mathematically, this can be expressed as:

• \( \log_{b}(a) + \log_{b}(c) = \log_{b}(ac) \)

In the original exercise, we applied this rule to combine \( \log_{2}(x+7) \) and \( \log_{2}(x) \), resulting in \( \log_{2}((x+7)x) = \log_{2}(x(x+7)) \). This simplifies complex logarithmic expressions and helps in solving equations by reducing the number of terms. The main trick is recognizing when and where to apply the rule to simplify the computation and interpretation of logarithmic equations.

A quadratic equation is a polynomial equation of degree two, which can be written in the standard form:

• \( ax^2 + bx + c = 0 \)

In the exercise, we transformed the logarithmic equation into a quadratic equation, \( x^2 + 7x - 8 = 0 \), by expressing it in exponential form. Solving quadratic equations typically involves the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides the solutions, or roots, of the quadratic equation by taking the coefficients \( a \), \( b \), and \( c \) from the equation and plugging them into the formula. For our case, solving yielded the potential solutions \( x = 1 \) and \( x = -8 \). Understanding how to efficiently solve quadratic equations is crucial in tackling various algebraic challenges.

To solve logarithmic equations effectively, converting them into exponential form is an essential step. This involves using the property:

• \( \log_{b}(y) = c \quad \text{is equivalent to} \quad y = b^c \)

In our problem, \( \log_{2}(x^2 + 7x) = 3 \) converts to the exponential equation \( x^2 + 7x = 2^3 \), simplifying to \( x^2 + 7x = 8 \). This approach allows us to bypass the logarithm, translating it into a more straightforward equation that can be tackled with algebraic techniques like solving quadratic equations. Mastering the conversion into exponential form is key to unlocking the solution to logarithmic equations.

Verifying the validity of solutions is critical when solving equations, especially those involving logarithms. Logarithms are only defined for positive values of their arguments. This means that any solution resulting in a logarithm of a negative number is not valid. After solving the quadratic equation, we found potential solutions \( x = 1 \) and \( x = -8 \).

• The value \( x = 1 \) is valid because it results in a positive argument for both \( x \) and \( x+7 \).
• In contrast, \( x = -8 \) would lead to negative arguments for \( x \) and \( x+7 \), invalidating it as a solution.

Always check the validity of solutions when dealing with logarithmic equations to ensure they meet all mathematical constraints.