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Chapter 3: Problem 12

In Exercises \(11-14,\) evaluate the logarithm using the change-of-base formula. Round your result to three decimal places. $$\log _{1 / 2} 4$$

Short Answer

Expert verified
The value of \(\log_{1/2}(4)\) when rounded to three decimal places is -2.000.

Step by step solution

01

Understand the problem

The problem is to evaluate \(\log _{\frac{1}{2}}(4)\) using the change-of-base formula, which can be given as \(\frac{log_c(a)}{log_c(b)}\). In this case, \(b = \frac{1}{2}\) and \(a = 4\).
02

Apply the change-of-base formula

To apply the change-of-base formula, simply replace \(a\) and \(b\) in the formula with 4 and \(\frac{1}{2}\) respectively. It doesn't matter much what base you change to, but usually base 10 or base \(e\) (which gives the natural logarithm), are the most common and convenient choices. Here, we'll use base 10: \(\frac{log_{10}(a)}{log_{10}(b)} = \frac{log_{10}(4)}{log_{10}(\frac{1}{2})}\)
03

Calculate the new log expressions

Now, calculate the values for \(log_{10}(4)\) and \(log_{10}(\frac{1}{2})\). These are approximately 0.602 and -0.301 respectively.
04

Perform the division

Finally, simply divide the two values: \(\frac{0.602}{-0.301}\).
05

Round off to three decimal places

Round off the final answer to three decimal places as instructed in the problem statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Logarithms
Logarithms are the mathematical concept used to solve what exponent is needed to get one number from another, using a specific base.
It is denoted as \(\log_b(a)\), where \(b\) is the base and \(a\) is the argument. For example, in the expression \(\log_2(8)\), it asks "to what power must we raise 2 to get 8?".
In this case, the answer is 3, because \(2^3 = 8\).
  • The base of a logarithm tells you what number is repeatedly multiplied.
  • The result of a logarithm tells you how many times the base is used.
  • Logarithms are the inverse operations of exponentials.
This means that understanding logarithms begins with understanding exponents.
Logarithms can be calculated in various bases, commonly base 10 (log) and base \(e\) (ln 鈥� natural logarithm). By grasping this concept, you can tackle various problems involving logarithms effortlessly.
The Role of Base Conversion in Logarithms
When working with logarithms, you may encounter various bases. To simplify calculations, converting the base of a logarithm can be advantageous.
The change-of-base formula is a tool that allows this conversion, and it's given by:
\[\log_b(a) = \frac{\log_c(a)}{\log_c(b)}\]
Here, the original base \(b\) is converted to the new base \(c\).
Common choices for \(c\) are base 10 or the natural logarithm base, \(e\).
  • This formula helps in computing logarithms using easily accessible calculators, primarily because they are often configured for base 10 or \(e\).
  • It also offers a formulaic approach to situations where a calculator might not be available or when working with theoretical data.
Utilizing base conversion makes evaluating logarithmic expressions manageable and intuitive, paving the way for better problem-solving techniques.
Calculating and Evaluating Logarithms
In mathematical evaluation, it is crucial to follow a structured approach to calculate and evaluate logarithmic expressions.
Using the change-of-base formula is one method. Take \(\log_{\frac{1}{2}}(4)\), for instance, which involves several steps:
  • Apply the change-of-base formula: Convert \( \log_{\frac{1}{2}}(4)\) to \( \frac{\log_{10}(4)}{\log_{10}(\frac{1}{2})} \).
  • Compute each log term separately: Find numerical values, \( \log_{10}(4) \approx 0.602 \) and \( \log_{10}(\frac{1}{2}) \approx -0.301 \).
  • Divide the results: \( \frac{0.602}{-0.301} = -2 \).
Careful calculation is essential here, using precise numbers or a calculator for accuracy.
Finally, remember to round your answer as required, like three decimal places in this example.
Mathematical evaluation through these steps aids in solving equations efficiently, ensuring both accuracy and comprehension.

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