Problem 16 - Precalculus with Calculus Previews

Question

In Problems \(13-18\), find the exact value of the given logarithm. $$ \log _{9} \frac{1}{3} $$

Step-by-Step Solution

Answer

The exact value is \(-\frac{1}{2}\).

Step 1: Understand the Logarithm Function

A logarithm \( \log_b(a) \) asks the question: "To what power must \( b \) be raised, to equal \( a \)?" In this problem, we are looking for \( \log_9 \left( \frac{1}{3} \right) \). This means: What power must \( 9 \) be raised to, to get \( \frac{1}{3} \)?

Step 2: Rewrite the Expression as an Equation

Let's set up the equation: \( 9^x = \frac{1}{3} \). Here, \( x \) is the power to which 9 must be raised to obtain \( \frac{1}{3} \).

Step 3: Express in Terms of Known Bases

We know that \( 9 = 3^2 \). So, the equation becomes \( (3^2)^x = \frac{1}{3} \). Simplifying the left side gives us \( 3^{2x} = \frac{1}{3} \).

Step 4: Convert the Fraction to a Negative Power

We know that \( \frac{1}{a} = a^{-1} \), so \( \frac{1}{3} = 3^{-1} \). Now, rewrite the equation as \( 3^{2x} = 3^{-1} \).

Step 5: Equate the Exponents

Since the bases are the same, we can set the exponents equal to each other: \( 2x = -1 \).

Step 6: Solve for x

Divide both sides of the equation \( 2x = -1 \) by 2 to get \( x = -\frac{1}{2} \). This is the solution to \( \log_9 \left( \frac{1}{3} \right) \).

Key Concepts

Logarithm Properties

Logarithms are mathematical operations that answer the question: "To what exponent must we raise a particular base to get another number?" This function has several useful properties:

• Product Rule: This states that the logarithm of a product is the sum of the logarithms, \[ \log_b(mn) = \log_b(m) + \log_b(n) \]
• Quotient Rule: This tells us that the logarithm of a quotient is the difference of the logarithms, \[ \log_b \left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \]
• Power Rule: According to this property, the logarithm of a power can be expressed as the exponent times the logarithm of the base, \[ \log_b(m^n) = n \cdot \log_b(m) \]
• Change of Base Rule: This property allows us to express a logarithm with one base in terms of logarithms with another base, \[ \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \]

Understanding these properties is crucial as they simplify complex logarithmic expressions. In the given exercise, recognizing how fractions and powers interact with logarithms is key—enabling us to solve problems more efficiently.

Exponential Equations

Exponential equations are those where variables appear in the exponent, such as in the expression \( 9^x = \frac{1}{3} \). To solve these equations, we need to understand how to manipulate bases and exponents effectively.

• One common strategy is to express both sides of the equation with the same base. This aligns the problem to more straightforward arithmetic, as shown in the exercise where 9 was expressed as \( 3^2 \).
• Once the bases are the same, the equation simplifies by equating the exponents, since \( a^m = a^n \) implies \( m = n \).
• The use of exponential equations is common in fields like finance for calculating compound interest, in biology for predicting population growth, and in physics for radioactive decay models.

Mastering exponential equations allows us to solve for unknown values in various real-world scenarios with precision.

Negative Exponents

Negative exponents represent the inverse of a positive exponent. The expression \( a^{-n} \) is equal to \( \frac{1}{a^n} \). This concept is crucial when dealing with fractions and simplifying expressions involves converting positive bases raised to negative powers.

• For instance, \( \frac{1}{3} \) can be rewritten as \( 3^{-1} \). This switch is fundamental in simplifying and equating exponents with like bases.
• Negative exponents often appear in scientific notation where we deal with very small numbers. For example, \( 10^{-3} \) represents 0.001.
• Understanding this concept is essential when manipulating algebraic expressions, especially in solving logarithmic and exponential equations where negative exponents naturally arise.

By solidifying the understanding of how negative exponents operate, students can solve a wider range of mathematical problems confidently and correctly.