Problem 95 - A Graphical Approach to Precalculus with Limits

Question

Use the change-of-base rule to find an approximation for each logarithm. $$\log _{29} 7.5$$

Step-by-Step Solution

Answer

\( \log_{29} 7.5 \approx 0.5983 \).

Step 1: Understand the Change-of-Base Formula

The change-of-base formula allows us to convert a logarithm in any base to a different base, typically base 10 or base e, which are easier to compute. The formula is:\[\log_b a = \frac{\log_c a}{\log_c b}\]where \(c\) is the new base we are converting to, commonly 10 or e (natural logarithm).

Step 2: Apply the Change-of-Base Formula

In this problem, we want to find \( \log_{29} 7.5 \). Using the change-of-base formula, we can convert it to:\[\log_{29} 7.5 = \frac{\log_{10} 7.5}{\log_{10} 29}\]Here, we chose base 10 (common logarithm) for simplicity, as it can be easily calculated using a calculator.

Step 3: Calculate Individual Logarithms

Now, we calculate the logarithms using base 10:1. \( \log_{10} 7.5 \) is approximately 0.8751.2. \( \log_{10} 29 \) is approximately 1.4624.

Step 4: Compute the Quotient

Now we plug these values back into the change-of-base formula:\[\log_{29} 7.5 = \frac{0.8751}{1.4624} \]Calculate the division to get the result.

Step 5: Final Calculation and Approximation

Performing the division \( \frac{0.8751}{1.4624} \) gives approximately 0.5983. Therefore, \( \log_{29} 7.5 \approx 0.5983 \).

Key Concepts

Logarithms

Logarithms are a mathematical concept used to describe the power or exponent that a base number must be raised to in order to obtain a given value. For example, in the logarithmic expression \( \log_b a \), \( b \) is the base, and the expression states "the power to which \( b \) must be raised to produce \( a \)."
Logarithms are the inverse of exponentiation. If \( b^c = a \), then \( \log_b a = c \). This inverse relationship makes logarithms particularly useful for simplifying expressions and solving equations that involve exponential growth or decay.
There are several types of logarithms based on the base used:

• Common logarithms (base 10) are represented as \( \log_{10} \).
• Natural logarithms (base \( e \), where \( e \approx 2.718 \)) are represented as \( \ln \).
• Binary logarithms (base 2) are useful in computer science.

Understanding how to work with different bases and utilizing the change-of-base formula helps in computing logarithms that might not be easily calculated with standard calculators.

Base Conversion

Base conversion is the process of changing a logarithm from one base to another. This is especially useful when dealing with logarithms of uncommon bases, which are not directly supported by most calculators.
The change-of-base formula is a powerful tool for this: \( \log_b a = \frac{\log_c a}{\log_c b} \). Here, you convert the original base \( b \) logarithm into an equivalent expression using a new base, \( c \). This allows evaluation using simpler, more-accessible bases like 10 or \( e \).
In practical terms, to convert \( \log_{29} 7.5 \) using base 10:

• Compute \( \log_{10} 7.5 \).
• Compute \( \log_{10} 29 \).
• Divide these values: \( \frac{\log_{10} 7.5}{\log_{10} 29} \).

This method simplifies complex calculations and provides a more user-friendly way to tackle exercises involving non-standard logarithm bases.

Mathematical Approximation

Mathematical approximation is crucial when working with values that aren't neat whole numbers or cannot be precisely calculated. Approximations allow for practical solutions in real-world applications where exact values are not necessary or possible.
In the context of logarithms, we often deal with irrational numbers that have non-terminating decimal expansions. Calculators typically provide decimal approximations for these, such as 0.8751 for \( \log_{10} 7.5 \) and 1.4624 for \( \log_{10} 29 \).
Though these approximations are not exact, they are sufficiently accurate for most purposes, ensuring that mathematical problems remain solvable without unnecessary complexity:

• Provides quick, practical solutions where exactitude isn't essential.
• Helps in simplifying complex mathematical expressions.
• Facilitates understanding and calculation of real-world problems.

Approximation skills enhance mathematical proficiency and encompass a broad range of applications from science to everyday problem solving.