Problem 19 - Algebra and Trigonometry

Question

Evaluate the expression. $$ \begin{array}{llll}{\text { (a) } \log _{3} 3} & {\text { (b) } \log _{3} 1} & {\text { (c) } \log _{3} 3^{2}}\end{array} $$

Step-by-Step Solution

Answer

(a) 1, (b) 0, (c) 2

Step 1: Evaluate \( \log_{3} 3 \)

To evaluate \( \log_{3} 3 \), we need to determine the power to which 3 must be raised to get 3. Since \(3^1 = 3\), it follows that \( \log_{3} 3 = 1 \).

Step 2: Evaluate \( \log_{3} 1 \)

To evaluate \( \log_{3} 1 \), we need to determine the power to which 3 must be raised to get 1. Since any number raised to the power of 0 is 1, \(3^0 = 1\). Thus, \( \log_{3} 1 = 0 \).

Step 3: Evaluate \( \log_{3} 3^2 \)

To evaluate \( \log_{3} 3^2 \), use the logarithmic identity \( \log_{b} b^a = a \). Here, \(b = 3\) and \(a = 2\), thus \( \log_{3} 3^2 = 2 \).

Key Concepts

Logarithmic Identities

Logarithms have special identities that simplify many calculations. These identities are shortcuts that help solve logarithmic expressions easily. One commonly used identity is the "Power Rule for Logarithms" which states that for any positive numbers \( b \) and \( a \), \( \log_b(b^a) = a \). This means when the base of the logarithm and the base of the exponent are the same, you can take the exponent out in front.
This identity was used in the step for evaluating \( \log_{3} 3^2 \). It tells us that if you have \( b^a \) and you take the log base \( b \), you get \( a \). This shortcut saves time and avoids complicated calculations. Another identity is \( \log_b(b) = 1 \) because the logarithm is asking the question, "To what power must \( b \) be raised to get \( b \)?" Obviously, that's 1.
Also, \( \log_b(1) = 0 \) because \( b^0 = 1 \). These identities make solving logarithmic problems simpler and quicker.

Logarithmic Equations

Logarithmic equations involve expressions where the logarithm of a number is set equal to something else. Learning how to manipulate these equations is essential for solving complex problems.
When solving logarithmic equations like those in the exercise, the fundamental step is to understand the base and the result you are trying to reach. For instance, in \( \log_{3} 3 = 1 \), you set the number 3 to the power of 1 to get 3, because in logarithms, you always want to find the power that gets you to the other number.

• A basic way to solve these equations is to exponentiate both sides to remove the logarithm.
• For example, if you have \( x = \log_b(y) \), this can be rewritten as \( b^x = y \) to eliminate the logarithm.

This process is helpful particularly when the logarithm isn’t as straightforward as in the examples provided. Practicing how to convert between exponents and logarithms will make these equations less intimidating.

Properties of Logarithms

The properties of logarithms are like tools in your math toolkit. These properties allow for the transformation and simplification of complex logarithmic expressions. One important property is the "Product Rule" which states \( \log_b(MN) = \log_b(M) + \log_b(N) \). This means if you are multiplying two numbers together and taking their log, you can break it into the sum of two logs.
Another useful property is the "Quotient Rule": \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \). It lets you turn a division inside the log into a subtraction outside of it. Lastly, we have the "Change of Base Formula," which is helpful when working with different bases. It shows \( \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \) where \( k \) is any positive number.

• These properties make logs far more flexible and help in solving equations more efficiently.
• Understanding these will allow you to handle a variety of problems with confidence and ease.

Mastering these properties will allow you to tackle logarithm problems like the pro you are becoming.