Problem 29
Question
Use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$ \log \sqrt{100 x} $$
Step-by-Step Solution
Verified Answer
The fully expanded form of the logarithmic expression is 1 + \(\frac{1}{2} \cdot \log(x)\)
1Step 1: Identify and Apply the Correct Logarithmic Property
First, we observe that the expression contains a square root; this is equivalent to raising it to the power of one half. We rewrite \( \sqrt{100x} \) as \( (100x)^{1/2} \). So, the initial equation is now: \( \log((100x)^{1/2})\).
2Step 2: Apply Logarithmic Properties
Next, we can apply the first logarithmic property: when the argument of a logarithm is raised to an exponent, we can bring that exponent out in front of the logarithm. The equation become: \( \frac{1}{2} \cdot \log(100x)\).
3Step 3: Solve Logarithmic Expression
Now, since the log of the product equals the sum of the log of each part, the equation becomes: \( \frac{1}{2} \cdot(\log(100) + \log(x))\). We know that log(100) to base 10 equals 2. So the equation becomes: \( \frac{1}{2} \cdot (2 + \log (x)) = 1 + \frac{1}{2} \cdot \log(x)\). We expanded the logarithmic expression as far as possible and evaluated part of it without a calculator
Key Concepts
Properties of LogarithmsExpanding LogarithmsEvaluating Logarithms
Properties of Logarithms
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.
Expanding Logarithms
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.
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.
Evaluating Logarithms
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.
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.
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