Problem 48 - Algebra 2 and Trigonometry

Question

In \(45-52,\) if \(\ln a=c,\) express each of the following in terms of \(c\) $$ \ln \frac{1}{a} $$

Step-by-Step Solution

Answer

\( \ln \frac{1}{a} = -c \)

Step 1: Identify the Given Expression

The problem asks us to express \( \ln \frac{1}{a} \) in terms of \( c \), given that \( \ln a = c \).

Step 2: Recall the Property of Logarithms

Recall the logarithmic property that states \( \ln \frac{1}{b} = -\ln b \). We can apply this property to our expression, where \( b = a \).

Step 3: Apply the Logarithmic Property

By applying the property from Step 2, we rewrite the expression: \( \ln \frac{1}{a} = -\ln a \).

Step 4: Substitute Known Values

Since \( \ln a = c \), substitute \( c \) into the expression: \( -\ln a = -c \).

Step 5: Conclusion

Thus, \( \ln \frac{1}{a} \) can be expressed as \( -c \).

Key Concepts

Understanding the Properties of Logarithms

Logarithms have specific properties that make them powerful tools for simplifying mathematical expressions. One crucial property of logarithms is the rule for taking the logarithm of a reciprocal. This property states that the logarithm of the reciprocal of a number is the negative logarithm of that number. In mathematical terms, this is expressed as: \( \ln \left( \frac{1}{b} \right) = -\ln b \).

This property works because logarithms are based on exponents, and taking the reciprocal of a number is equivalent to applying a negative exponent.

• Recall that \( b^{-1} = \frac{1}{b} \).
• In logarithmic terms, \( \ln b^{-1} = -\ln b \).

This property is heavily utilized to transform complex logarithmic expressions into simpler forms, making solving problems easier. Like in the exercise, once you know \( \ln a = c \), you can quickly derive that \( \ln \frac{1}{a} = -c \) using this straightforward property.

Exploring the Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a special type of logarithm that uses the base \( e \), where \( e \) is approximately equal to 2.71828. The natural logarithm is widely used in mathematics, science, and engineering because of its natural occurrence in growth processes, particularly exponential growth and decay.

Natural logarithms have key relations and uses, including:

• The inverse relationship with exponentials: \( \ln(e^x) = x \) and \( e^{\ln x} = x \).
• They help in solving equations involving exponential and logarithmic expressions.
• The derivative of a natural logarithm function \( \ln(x) \) is \( \frac{1}{x} \), making it crucial in calculus.

Being comfortable with natural logarithms is important in both academic and real-world applications because they are a foundational concept in higher mathematics.

Solving Logarithmic Expressions

Logarithmic expressions often show up in various mathematical contexts, requiring simplification or solving for unknowns. These expressions can be simplified using properties of logarithms or by rewriting them in terms of known values.

When given variables, as in the exercise, the key steps to tackle logarithmic expressions include:

• Identifying any logarithmic properties applicable to the expression, like the reciprocal property or the product, quotient, and power rules.
• Substituting known values into the expression to simplify or solve for a variable. For instance, knowing \( \ln a = c \) allows for quick substitution to express \( \ln \frac{1}{a} \).
• Simplifying the expression using algebraic manipulation: converting a problem into a manageable form by using negative signs, exponents, or other algebraic steps.

The ease of working with logarithmic expressions comes from practice and familiarity with these rules, which turn initially complex problems into straightforward calculations.