Problem 18 - Algebra A Combined Function

Question

Find the exact value of each logarithm. $$ \log \frac{1}{10} $$

Step-by-Step Solution

Answer

The exact value is -1.

Step 1: Understand the Property of Logarithms

Recall the property that \( \log_b\left(\frac{1}{a}\right) = -\log_b(a)\). This property states that the logarithm of a reciprocal is the negative logarithm of the base.

Step 2: Apply the Property to the Problem

Notice that \( \frac{1}{10} \) is the reciprocal of 10. Using the property from Step 1, we can write \[ \log\left(\frac{1}{10}\right) = -\log(10). \]

Step 3: Solve Using Basic Logarithmic Identity

We know that \( \log(10)\) in base 10 is 1, because 10 is the base of the common logarithm and \( 10^1 = 10. \). So,\[ \log(10) = 1. \]

Step 4: Substitute and Simplify

Substitute the value from Step 3 into the equation from Step 2:\[ -\log(10) = -1. \]

Step 5: Final Answer

Therefore, the exact value of \[ \log \left(\frac{1}{10}\right) \] is \(-1\).

Key Concepts

Logarithmic Properties

Logarithmic properties are fundamental rules that help simplify and understand logarithms. Logarithms are the inverse operations of exponentials, which means they allow us to find the power to which a number (the base) must be raised to obtain another number. Understanding these properties can make complex calculations much simpler.
One key property used in our exercise is the property of the reciprocal. This property states \[\log_b\left(\frac{1}{a}\right) = -\log_b(a)\]This means that taking the logarithm of a reciprocal is equivalent to taking the negative of the logarithm of that number. It's useful in both theoretical and practical math problems.
Whenever you encounter a logarithm of a fraction like \(\log(\frac{1}{a})\), this property is an easy way to convert it into a more straightforward calculation. Understanding such properties allows you to tackle logarithms with confidence.

Reciprocal

The reciprocal of a number is simply one divided by that number. For example, the reciprocal of 10 is \(\frac{1}{10}\). Reciprocals play a critical role in various mathematical operations and concepts, including logarithms.
When we discuss the reciprocal in the context of logarithms, we're often interested in how the change in the expression mirrors a change in the logarithmic value. The logarithmic property of reciprocals tells us that the logarithm of a reciprocal results in a negative logarithm of the original number. This insight is crucial when handling logarithmic equations and is directly applied in the exercise we're exploring.
Remember: whenever you see a reciprocal, it's a sign that the logarithmic value will change its sign to negative.

Common Logarithm

A common logarithm is a logarithm with a base of 10, often written simply as \(\log(x)\). It's called "common" because base 10 is the numeric base most frequently used in applications and calculations.

• Common logarithms are particularly handy in fields like science and engineering where exponential growth or decay is observed frequently.
• They simplify calculations since our numeric system is base 10.

In solving the given exercise, we used the knowledge that \(\log(10) = 1\), reflecting the property that the base raised to the power of the logarithm equals the number (\(10^1 = 10\)). Recognizing when a logarithm is in base 10 allows for quick solutions, making problems easier and less time-consuming.

Base 10

The concept of a base in logarithms refers to the number that is raised to a power to achieve another number. Base 10 is commonly used because it matches our decimal system. In the context of logarithms, when we refer to \(\log(x)\), and there is no base indicated, it is understood to be base 10.
Base 10 logarithms are intuitive when dealing with powers and exponential growth. They are essential for solving problems involving exponential equations, such as those found in financial calculations, scientific data analysis, and engineering problems.

• Base 10 is often referred to as the "common base" in logarithmic terms.
• Dealing with base 10 helps standardize calculations across various fields of study.

In our specific exercise, understanding that we were dealing with base 10 allowed us to simply apply the logarithm rules and quickly find out that \(\log(10) = 1\), leading us to find \(\log(\frac{1}{10}) = -1\). This understanding simplifies many mathematical problems by reducing the complexity of calculations.