Problem 76 - College Algebra with Modeling and Visualization

Question

Use the change of base formula to approximate the logarithm to the nearest thousandth. $$ \frac{\log _{7} 125}{\log _{7} 25} $$

Step-by-Step Solution

Answer

The logarithm is approximately 1.500.

Step 1: Understanding the Change of Base Formula

The change of base formula is used to convert a logarithm of any base to a quotient of logarithms in any other base. It states that \( \log_b a = \frac{\log_k a}{\log_k b} \), where \( k \) is a new base, commonly 10 or \( e \).

Step 2: Apply the Change of Base Formula

For \( \log_7 125 \), apply the change of base formula to express it in terms of base 10: \( \log_7 125 = \frac{\log_{10} 125}{\log_{10} 7} \). Similarly, \( \log_7 25 = \frac{\log_{10} 25}{\log_{10} 7} \).

Step 3: Simplifying the Expression

Substitute the expressions from Step 2 into the original fraction: \( \frac{\log_7 125}{\log_7 25} = \frac{\frac{\log_{10} 125}{\log_{10} 7}}{\frac{\log_{10} 25}{\log_{10} 7}} \).

Step 4: Canceling Terms

Notice that \( \log_{10} 7 \) appears in both the numerator and the denominator, cancel them out: \( \frac{\log_{10} 125}{\log_{10} 25} \).

Step 5: Calculate the Logs

Use a calculator to find \( \log_{10} 125 \approx 2.0969 \) and \( \log_{10} 25 \approx 1.3979 \).

Step 6: Compute the Quotient

Divide the two results obtained in the previous step: \( \frac{2.0969}{1.3979} \approx 1.500 \), rounded to the nearest thousandth.

Key Concepts

Logarithms

Logarithms are a way to express exponential relationships. Think of them as the opposite of exponentiation. In simple terms, if you have an equation like \( b^x = a \), the logarithm lets you solve it by finding \( x \). This looks like \( \log_b a = x \), meaning what power times itself gives \( a \).

Here are some key points about logarithms to consider:

• The **base** of a logarithm is what you're multiplying.
• **Logarithms can be of any positive base except 1** – common ones are 10 (common log) and \( e \) (natural log).
• **Converting between bases** allows the use of familiar bases, making calculations easier.
• **Properties to note** include the product, quotient, and power rules of logarithms, which simplify complex expressions.

Understanding logarithms is crucial for working with various scientific and engineering problems as they easily scale up or down large numbers.

Base Conversion

The concept of base conversion in logarithms helps in shifting a problem to a more convenient format. You might often need to switch logarithmic bases to simplify calculations, especially on a calculator that supports only base 10 or with natural logs. The Change of Base Formula comes in handy here and is given by: \[\log_b a = \frac{\log_k a}{\log_k b}\]Here, you can choose any base \( k \) to convert to, but standard practice opts for base 10 or \( e \, (\log_{10}) \).

When applying this formula, keep in mind:

• **Ease of calculation**: Base 10 simplifies manual calculations using a calculator's common log function.
• **Universal application**: Whether for small or large numbers, the formula works consistently across different scales.
• In this exercise, converting from base 7 to base 10 enables easy computation of \( \log_7 125 \) and \( \log_7 25 \) using a calculator.

Base conversion ensures that working out complicated logarithms is manageable and accessible.

Mathematical Approximation

Mathematical approximation is a critical tool in numerical computation. When we use logarithms, sometimes the resulting values don't fit neatly into whole numbers. Calculators often help us reach a precise decimal value, as seen in the exercise using these values:

• \( \log_{10} 125 \approx 2.0969 \)
• \( \log_{10} 25 \approx 1.3979 \)

For precise results like rounding, knowing how and when to approximate is crucial. Here are some steps to focus on:

• **Accurate Decimal Placement**: Always decide beforehand how much precision you need, here it's to the nearest thousandth.
• **Numerical Rounding**: Apply rules systematically, such as rounding \( \frac{2.0969}{1.3979} \approx 1.500 \).
• **Practical Implications**: Approximated outcomes are sufficient for many practical applications like engineering and physics.

Understanding approximations ensures you arrive at functional and relevant solutions without unnecessary complexity.