Problem 4 - College Algebra with Modeling and Visualization

Question

Evaluate each expression by hand, if possible. (a) \(\log 10,000\) (b) \(\log (-\pi)\) (c) \(\log \sqrt{0,001}\) (d) \(\log 8^{\circ}\)

Step-by-Step Solution

Answer

(a) 4, (b) Undefined, (c) -1.5, (d) Not evaluable due to unclear notation.

Step 1: Evaluate \( \log 10,000 \)

The logarithm with base 10, noted as \( \log \), asks for the power to which 10 must be raised in order to get 10,000. Since \( 10^4 = 10,000 \), it follows logically that \( \log 10,000 = 4 \).

Step 2: Evaluate \( \log (-\pi) \)

The logarithm function \( \log x \) for base 10 is only defined for positive values of \( x \). Since \( -\pi \) is a negative number, \( \log (-\pi) \) is not defined.

Step 3: Evaluate \( \log \sqrt{0.001} \)

First, express \( \sqrt{0.001} \) as an exponent: \( \sqrt{0.001} = (0.001)^{1/2} \). Then use the property \( \log(a^b) = b \cdot \log a \). Since \( 0.001 = 10^{-3} \), we have \( \sqrt{0.001} = 10^{-1.5} \). Therefore, \( \log \sqrt{0.001} = -1.5 \).

Step 4: Evaluate \( \log 8^{\circ} \)

"\( 8^{\circ} \)" typically represents a degree notation and not a number. Therefore, \( \log 8^{\circ} \) is not an expression that can be evaluated in the mathematical sense. In context, it is possibly meant to find \( \log 8 \), which results in decimal value \( \log 8 \approx 0.903 \), but with unclear notation, this computation might not be intended.

Key Concepts

Exponents

Exponents are a fundamental mathematical concept where a number, called the base, is multiplied by itself a number of times determined by the exponent. For example, in the expression \(10^4\), the base is 10 and the exponent is 4. This means that 10 is multiplied by itself four times: \(10 \times 10 \times 10 \times 10 = 10,000\).
This concept is crucial for understanding logarithms because logarithms can be seen as the inverse operation of exponentiation. Exponents are used to express numbers in a simplified and more manageable way, especially when dealing with very large or very small numbers.

• Positive exponents indicate how many times to use the base as a factor.
• Negative exponents indicate the reciprocal, or the inverse, of the base being used as a factor a certain number of times.
• Fractional exponents denote roots, where the numerator is the power and the denominator is the root.

In the exercise, the concept of exponents is evident when expressing 0.001 as \(10^{-3}\), and \(\sqrt{0.001}\) as \(\sqrt{10^{-3}} = 10^{-1.5}\). This property helps us evaluate logarithmic expressions more easily.

Base 10 logarithm

Base 10 logarithms, also known as common logarithms, use 10 as their base. They are often denoted as \(\log\) and are commonly used in fields like engineering and the sciences. A logarithm answers the question: "To what power must the base (10) be raised to produce a given number?" For instance, \(\log 100 = 2\) because \(10^2 = 100\).
The base 10 logarithm is particularly convenient for calculations involving powers of ten, which aligns well with our decimal numeral system.

• The logarithm of a power of ten is simply the exponent (e.g., \(\log 10^4 = 4\)).
• A logarithm is undefined for zero and negative numbers because you cannot raise a positive number to any real power to obtain a negative result or zero.

In our exercise, evaluating \(\log 10,000 = 4\) shows this, as 10 must be raised to the 4th power to equal 10,000. Additionally, \(\log (-\pi)\) is not defined, illustrating how negative numbers cannot be used in base 10 logarithms.

Evaluation of expressions

Evaluating expressions, particularly logarithmic expressions, involves using mathematical rules and understanding concepts like exponents and properties of logarithms.
For example, logarithmic properties that help in evaluating expressions include:

• \(\log(a^b) = b \cdot \log a\), which allows the exponent to be brought in front as a multiplier.
• \(\log(abc) = \log a + \log b + \log c\), which breaks down the logarithm of a product into a sum.
• \(\log \frac{a}{b} = \log a - \log b\), which transforms the logarithm of a quotient into a difference.

Comprehending these properties streamlines the evaluation process, turning complex expressions into simpler calculations.
In the exercise, evaluating \(\log \sqrt{0.001}\) required converting the square root into a power (\(\sqrt{0.001} = 10^{-1.5}\)), allowing us to use the rule \(\log(a^b) = b \cdot \log a\) to find a manageable solution.