Problem 20 - Algebra and Trigonometry

Question

\(15-24\) Evaluate the expression. $$ \begin{array}{llll}{\text { (a) } \log _{5} 125} & {\text { (b) } \log _{49} 7} & {\text { (c) } \log _{9} \sqrt{3}}\end{array} $$

Step-by-Step Solution

Answer

(a) 3, (b) 1/2, (c) 1/4

Step 1: Understanding the problem

We need to evaluate logarithms for each of the expressions given in parts (a), (b), and (c). Logarithms are the inverse operations of exponentiation, meaning if \(a^b = c\), then \(\log_a c = b\). We will use this understanding to solve the problems.

Step 2: Evaluating \(\log_5 125\)

For part (a), we need to find \(\log_5 125\). We rewrite 125 as a power of 5. We know that \(5^3 = 125\), so \(\log_5 125 = 3\).

Step 3: Evaluating \(\log_{49} 7\)

For part (b), we need to evaluate \(\log_{49} 7\). We recognize that 49 is essentially \(7^2\). Therefore, we have \(\log_{49} 7 = \frac{1}{2}\), because 7 is the square root of 49.

Step 4: Evaluating \(\log_9 \sqrt{3}\)

For part (c), we need to find \(\log_9 \sqrt{3}\). We can express \(\sqrt{3}\) as \(3^{\frac{1}{2}}\) and recognize 9 as \(3^2\). Therefore, using the change of base property, \(\log_9 \sqrt{3} = \log_{3^2} 3^{\frac{1}{2}} = \frac{1}{4}\) because the base \(3^2\) exponent needs to be multiplied by \(\frac{1}{2}\) to get \(3\).

Key Concepts

Inverse Operations

When we talk about inverse operations, we refer to two mathematical processes that undo each other. Logarithms and exponentiation are classic examples of inverse operations. If you know that an operation can be reversed, it gives you a powerful tool to solve equations. For instance, exponentiation involves raising a number to a power, while a logarithm helps you find that power. If you have a number expressed as a power of another, the logarithm can tell you what that power is. So, with multiplication and division, addition and subtraction, logarithms and exponentiation also pair together as inverse operations.

Exponentiation

Exponentiation is a mathematical operation involving a base and an exponent. In simple terms, it refers to multiplying a base number by itself a specified number of times. For example:

• If you have the expression \(5^3\), it means you multiply 5 by itself 3 times: \(5 \times 5 \times 5 = 125\).
• This calculation shows up frequently when working with logarithms, as it helps transform numbers into their power bases.

The relationship between exponentiation and logarithms becomes clear when you understand that if \(a^b = c\), then \(b\) is the logarithm of \(c\) with base \(a\), or \(\log_{a} c = b\). This connection is crucial for evaluating and solving logarithmic equations effectively.

Change of Base Property

The change of base property is an important tool that allows us to evaluate logarithms with different bases. It states that you can rewrite any logarithm in terms of logs with a different base by using a specific formula.To use the change of base property, use the formula:\[\log_b a = \frac{\log_c a}{\log_c b}\]Where \(c\) is any positive number you choose, often 10 or \(e\) (the base of natural logarithms). For example, evaluating \(\log_{9}\sqrt{3}\) can be thought of using the base 3 since 9 can be written as \(3^2\) and \(\sqrt{3}\) is \(3^{\frac{1}{2}}\). Using the change of base property simplifies this process, particularly when performing calculations manually.

Logarithmic Equations

Logarithmic equations are equations that involve logarithms and can sometimes seem tricky at first. However, by understanding their relationship with exponents, solving them becomes more manageable. When you see that an equation involves a log, consider how you can "flip" it to look like an exponent equation.To solve a logarithmic equation, ensure that you:

• Identify whether rewriting in an exponential form helps.
• Use properties of logarithms like product, quotient, and power rules if needed.

When dealing with an equation like \(\log_5 125 = x\), ask yourself: "What power do I raise 5 to, so I get 125?" This understanding makes it equal to \(5^x = 125\). By familiarizing yourself with basic log rules and the concept of inverse operations, even seemingly complex logarithmic equations become far more approachable.