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Understanding the properties of logarithms is a crucial step in solving logarithmic equations effectively. In our exercise, these properties helped us simplify and manipulate the equation.One property we used is the power rule, which states that multiplying a logarithm by a number is equivalent to the logarithm of the power of the number. Formally, it's expressed as:

• \(a \log b = \log b^a\)

Another important property is the subtraction rule, which allows us to simplify the difference of two logarithms to a single logarithm. It's expressed as:

• \(\log a - \log b = \log\left(\frac{a}{b}\right)\)

By using these properties, the original equation was condensed from multiple logarithmic expressions on the right side to a single expression, making it easier to work with.

After manipulating the logarithmic equation, we ended up with a quadratic equation. A quadratic equation is a second-degree polynomial equation of the form:

• \(ax^2 + bx + c = 0\)

In our solved exercise, the quadratic equation took the form:

• \(y^2 - 8y + 7 = 0\)

These kinds of equations often have two solutions. They are typically solved by factoring, using the quadratic formula, or by completing the square. Understanding this concept is essential because it's a common form of an equation you will encounter in mathematics.

Factoring is a method used to solve quadratic equations by expressing them as a product of two binomials. In essence, an equation like \(y^2 - 8y + 7 = 0\) can be rewritten using factoring as:

• \((y - 7)(y - 1) = 0\)

Through factoring, you can easily find the solutions to the quadratic equation. The factors provide the values of \(y\) where each binomial equals zero:

• If \((y - 7) = 0\), then \(y = 7\)
• If \((y - 1) = 0\), then \(y = 1\)

These solutions arise because the product of the factors equals zero only when at least one of the factors is zero. Factoring is not just a tool for solving equations but also helps in simplifying expressions in algebra.

Once we solve an equation, especially a transformed one, verifying the solutions is crucial to ensure they satisfy the original equation.In this exercise, after finding the possible solutions \(y = 7\) and \(y = 1\), we substitute them back into the original logarithmic equation. This confirms their validity:

• For \(y = 7\), substituting gives \(\log(50) = \log(50)\)
• For \(y = 1\), substituting gives \(\log(8) = \log(8)\)

Both substitutions hold true, confirming both solutions are correct. Verification helps eliminate errors that might arise during manipulation and ensures that the solutions conform to the original problem.