91Ó°ÊÓ

To expand and simplify logarithmic expressions, understanding the properties of logarithms is essential. In this exercise, several properties are used effectively.

• Power Rule: If you have a term raised to a power inside a logarithm, you can bring that power to the front as a multiplier. This is expressed as \( \log_b(a^n) = n\log_b(a) \). In our solution, \( (100x)^{1/2} \) was rewritten to utilize this rule.
• Product Rule: For the product of two numbers inside a logarithm, you can express it as the sum of two separate logarithms: \( \log_b(a \cdot c) = \log_b(a) + \log_b(c) \). In the exercise, \( \log(100x) \) was expanded using this rule.
• Log Base Rule: Finally, being familiar with common logarithmic values is crucial. For instance, knowing that \( \log_{10}(100) = 2 \) simplifies evaluation.

Grasping these properties allows simplification of even complex expressions with ease, as demonstrated.

When dealing with logarithmic expressions, expansion simplifies them into more manageable parts. For our specific example, this means transforming the combined term within, \( \sqrt{100x} \), into different elements.

The first step involves rewriting the square root as a fractional exponent: \( \sqrt{100x} = (100x)^{1/2} \). By employing the logarithmic properties, we pull out the exponent, transforming it into \( \frac{1}{2} \log(100x) \).

Next, we use the product rule to break down the logarithm of a product \( \log(100x) \) into separate logarithms: \( \log(100) + \log(x) \). Thus the expression expands to \( \frac{1}{2}(\log(100) + \log(x)) \), making each part ready for further evaluation.

This step-by-step expansion aids in isolating and evaluating components, allowing us to approach finding the answer methodically.

Evaluation of logarithmic expressions helps attain specific numerical values, much like simplification. In the exercise, we evaluated \( \log(100) \), which required recognizing common log values: \( \log_{10}(100) = 2 \).

Recognizing benchmark values such as \( \log_{10}(10) = 1 \) or \( \log_{10}(1000) = 3 \) allows swift evaluation without relying on calculators. In this case, knowing that \( 100 = 10^2 \) makes it easy to compute \( \log_{10}(100) = 2 \), leveraging the power rule.

Combining this evaluated part with others gives \( \frac{1}{2}(2 + \log(x)) = 1 + \frac{1}{2}\log(x) \), a simpler, comprehensible expression. Such comprehension powers not only problem-solving ability but also confidence in handling complex logarithmic challenges.