91Ó°ÊÓ

Logarithms can be a bit tricky, but once you get the hang of it, they can simplify complex equations with ease. A crucial part of simplifying logarithmic expressions is using rules like the product rule. The product rule for logarithms states that the logarithm of a product is equal to the sum of the logarithms of its factors. In mathematical terms, this is expressed as:

• \( \log_b (MN) = \log_b M + \log_b N \)

In our original exercise, we had \( \log 2 + \log (3x - 4) \) on the right side of the equation. By applying the product rule here, the sum of these two logs simplified to a single logarithm: \( \log(2(3x - 4)) \), which further becomes \( \log(6x - 8) \). This is an excellent example of how the product rule can transform a logarithmic equation, making it easier to handle.

The power rule is another essential tool when solving logarithmic equations. This rule indicates that a coefficient before a logarithm can be turned into an exponent, allowing further simplification. Mathematically, the power rule is stated as:

• \( b \log_b a = \log_b a^b \)

In our example, we applied the power rule to the expression \( 2 \log x \). According to the power rule, we transformed it to \( \log x^2 \). This manipulation allows us to write exponents just like any other multiplication and vaults our problem-solving to a logarithmic equation that parallels the left and right side easily. Once applied, the power rule creates the opportunity to remove logarithms and proceed with solving the equation easily after simplification.

The final step in many exercises involving logarithms, including our example, often leads us to a quadratic equation. Understanding how to solve quadratic equations is crucial. A quadratic equation typically looks like:

• \( ax^2 + bx + c = 0 \)

For our exercise, after removing the logarithms, we had \( x^2 = 6x - 8 \). Rearranging it gave us the standard form \( x^2 - 6x + 8 = 0 \). To solve a quadratic equation like this, you might use factoring, the quadratic formula, or even graphing. Here, we factored the equation into \((x - 2)(x - 4) = 0\). This equation implies that the solutions for \( x \) are the values that make either \( x - 2 = 0 \) or \( x - 4 = 0 \). Hence, the solutions are \( x = 2 \) or \( x = 4 \). Both solutions must be verified by substituting them back into the original equation to ensure they satisfy it.