Problem 38 - College Algebra with Corequisite Support

Question

For the following exercises, use the definition of common and natural logarithms to simplify. \(2 \log (.0001)\)

Step-by-Step Solution

Answer

The simplified expression is \(-8\).

Step 1: Understand the Expression

The given expression is \(2 \log (0.0001)\). Here, \(\log\) represents the logarithm with base 10, which is the common logarithm. Our goal is to simplify this expression using properties of logarithms.

Step 2: Use the Logarithm Property

Use the property of logarithms that states \(a \log_b (x) = \log_b (x^a)\). Applying this to our expression, we have \(2 \log (0.0001) = \log (0.0001^2)\).

Step 3: Simplify the Exponentiation Inside the Logarithm

Calculate \((0.0001)^2\). This is equal to \(0.00000001\) because when you square a power of 10, you double the exponent: \((10^{-4})^2 = 10^{-8}\). Therefore, \(0.0001^2 = 10^{-8}\), so we have \(\log (0.00000001)\).

Step 4: Apply the Definition of Logarithm

Use the definition of the logarithm: \(\log_{10} (10^{-8}) = -8\) because the logarithm asks "To what power must 10 be raised to obtain \(10^{-8}\)?". Since \(10^{-8} = 10^{-8}\), the logarithm is \(-8\).

Key Concepts

Common Logarithm

The common logarithm is a logarithm that uses base 10. It is often denoted simply as \( \log \) without a subscript. This is because base-10 logarithms are common in scientific and engineering calculations. The common logarithm answers the question, "To what power must 10 be raised to get a specific number?"
For example:

• \( \log 100 = 2 \) because \( 10^2 = 100 \)
• \( \log 1000 = 3 \) because \( 10^3 = 1000 \)

Understanding this concept is crucial for simplifying expressions involving common logarithms, as in the exercise \(2 \log (0.0001)\). By rewriting the problem in terms of powers of 10, you make the calculations straightforward and manageable.

Properties of Logarithms

Logarithms have several key properties that are incredibly useful for simplifying complex expressions. These properties are based on the idea that logarithms are exponents.

• Power Property: \( a \log_b (x) = \log_b (x^a) \). This means that if you have a coefficient in front of a logarithm, you can move it as an exponent of the argument inside the log.
• Product Property: \( \log_b (xy) = \log_b (x) + \log_b (y) \). Multiply two numbers and take the log, or take the log of each and add them.
• Quotient Property: \( \log_b \left( \frac{x}{y} \right) = \log_b (x) - \log_b (y) \). Take the log of a fraction as the difference of the logs.

In the exercise, we used the power property to simplify \( 2 \log (0.0001) = \log (0.0001^2) \). Utilizing these properties allows for transforming logarithmic expressions efficiently.

Exponentiation

Exponentiation is the mathematical operation of raising a number, called the base, to the power of an exponent. In the context of logarithms, it is often necessary to reverse the log operation by exponentiating to understand what's happening.Consider the expression \( x^n \), where \( x \) is the base and \( n \) is the exponent. For example, \( (10^{-4})^2 = 10^{-8} \). This is a core part of the step-by-step solution where \( 0.0001^2 = 10^{-8} \) was needed to simplify inside the logarithm.

Understanding how powers and exponents work makes it simpler to handle expressions like \( 2 \log (0.0001) \). When squaring \( 0.0001 \), knowing the rules of exponentiation helps transition to the simplified expression inside the logarithm.

Definition of Logarithms

Logarithms are the inverse operations of exponentiation. If you have \( b^y = x \), then the logarithm base \( b \) of \( x \) is \( y \). This operation measures how many times the base \( b \) must be multiplied by itself to achieve \( x \).Let's connect this with an example:

• If \( 10^{-8} = 10^{-8} \), then \( \log_{10}(10^{-8}) = -8 \).

This simple identity focuses on the question: "What power of 10 equals \( 10^{-8} \)?" Understanding this fundamental definition of logarithms is key when simplifying logarithmic expressions like in the exercise, ensuring that you interpret the expressions accurately.