Problem 32 - Precalculus : Building Concepts and Connections

Question

Evaluate each expression without using a calculator. $$\log _{1 / 3} 9$$

Step-by-Step Solution

Answer

The value of \(\log _{1 / 3} 9\) is \(2\).

Step 1: Express as powers of 3

First, express the base \(1/3\) and the number \(9\) as powers of \(3\). \(1/3\) equals \(3^{-1}\) and \(9\) equals \(3^2\). So \(\log _{1 / 3} 9\) can be written as \(\log _{3^{-1}} {3^2}\).

Step 2: Use Logarithm power rule

The power rule of logarithm states that \(\log _{b} {a^n}\) equals \(n \cdot \log _{b} {a}\). Apply this rule to the expression \(\log _{3^{-1}} {3^2}\) to get: \(2 \cdot \log _{3^{-1}} {3}\).

Step 3: Simplify Logarithm of base equals to itself

A number's logarithm at its own base is always \(1\). This means that \(\log _{3^{-1}} {3}\) is \(1\). Replace \(\log _{3^{-1}} {3}\) with \(1\) to get: \(2 \cdot 1\).

Step 4: Final Simplification

Now, perform the arithmetic: \(2 \cdot 1\) equals \(2\).

Key Concepts

Powers of 3

When working with logarithms, expressing numbers as powers can simplify the calculation process. In this exercise, we need to evaluate \(\log _{1 / 3} 9\). To do this efficiently, we first express both \(1/3\) and \(9\) as powers of 3.

• \(1/3\) can be rewritten as \(3^{-1}\) because dividing by 3 is the same as multiplying by its reciprocal, which is \(3^{-1}\).
• Similarly, \(9\) can be expressed as \(3^2\) because \(9\) is the product of two 3s: \(3 \times 3\).

By expressing numbers in this way, we align our terms with bases as powers of the same number (in this case, 3), laying the groundwork for the next steps in solving the logarithm problem. This approach helps to seamlessly apply logarithmic rules.

Logarithm Power Rule

The logarithm power rule is a convenient tool when working with logarithmic expressions. This rule states that \(\log_{b} a^n = n \cdot \log_{b} a\). Let's explore how this rule helps simplify expressions.

• In our case, the expression \(\log _{3^{-1}} {3^2}\) fits the format \(\log_{b} a^n\) where \(b = 3^{-1}\) and \(a = 3\).


• By applying the power rule, we transform the expression to \(2 \cdot \log _{3^{-1}} {3}\).

Notice how the exponent (which is 2 here) moves out in front, multiplying the entire logarithmic expression. By using the power rule, we simplify complex exponential expressions inside a logarithm, which makes the following steps much more manageable. Understanding and applying this rule is crucial in logarithmic simplifications and evaluations.

Simplifying Logarithms

Simplifying logarithmic expressions is often the final step in solving them, and it's crucial for reaching a numerical answer. Here, understanding a key property of logarithms helps us simplify \(\log _{3^{-1}} {3}\).

• One essential property is that \(\log_{b}{b} = 1\) for any positive base \(b\). This means that the logarithm of a number at its own base is always 1.


• Fortunately, this property extends to reciprocals such as \(3^{-1}\). Therefore, \(\log _{3^{-1}} {3}\) simplifies to \(-1\) because it's essentially \(\log_{b^{-1}}{b}\), meaning considering the inverse base.
• Therefore, when we calculate the expression \(2 \cdot (-1)\), the result is \(-2\).

With these simplifications, we've managed to deduce the result efficiently. Understanding these properties and simplification techniques is key to handling more complex logarithmic problems. This ensures you're equipped to tackle challenges involving logarithms confidently.