91Ó°ÊÓ

Logarithms have several important properties that make solving equations much simpler. One key property is the product rule, which states that

• \(\log_a(bc) = \log_a b + \log_a c\)

This property allows us to combine logs by multiplying the arguments. For example, in the given exercise, \(\log(x-2) + \log 5\) becomes \(\log[(x-2) \times 5]\).
Another essential property is the power rule,

• \(\log_a(b^n) = n \times \log_a b\)

These rules are crucial when rearranging and simplifying logarithmic expressions, making it easier to find solutions. By using the properties, we simplify complex logarithmic expressions into more manageable forms.

Understanding the domain of logarithmic expressions is crucial when working through equations. The domain is the set of all possible values that the variable can take without causing the expression to be undefined. For a logarithm \(\log_a x\), the domain consists of all positive real numbers because the argument \(x\) must be greater than zero.
In the exercise, \(\log(x-2)\) suggests that \(x-2\) has to be greater than zero. Thus, \(x > 2\). This restriction ensures no negative or zero values are inside the logarithm, which would make it undefined.
Before solving a logarithmic equation, checking the domain helps avoid accepting solutions that are not valid within the context of the problem.

When solving logarithmic equations, converting them to a non-logarithmic form is often a necessary step. Once we have simplified the equation using properties of logarithms, we equate the arguments of the logarithms when both sides have the same base. For example, the equation \(\log[(x-2)5] = \log 100\) leads to the conclusion that

• \((x-2) \times 5 = 100\)

With the logs removed, this becomes a basic algebra problem.
Solve it step-by-step:

• Divide both sides by 5: \(x - 2 = 20\)
• Add 2 to both sides: \(x = 22\)

This methodical approach to solving equations ensures accuracy and clarity in finding the solution.

After determining the exact solution to a logarithmic equation, a decimal approximation might be necessary. This is especially useful when the problem specifically asks for a solution to a certain number of decimal places, such as two decimal places.
In this particular exercise, once you have determined the exact value of \(x = 22\), you would check if further approximation is needed.
A calculator can be used to ensure accuracy of decimal places. Decimal approximations make complicated numbers more manageable, especially in practical scenarios or further calculations. Although \(x = 22\) doesn't require approximation, understanding the process and when it's needed is a valuable skill.