Problem 100 - Elementary and Intermediate Algebra

Question

Use the change-of-base formula to find logarithm to four decimal places. \(\log _{5} 10\)

Step-by-Step Solution

Answer

\(\log_5 10 \approx 1.4307\)

Step 1: Understand the Change of Base Formula

The change of base formula for logarithms allows us to convert a logarithm to a different base using the formula: \[ \log_b a = \frac{\log_c a}{\log_c b} \] where \(c\) is a new base. Commonly, we use base 10 or base \(e\) (natural logarithm, \(\ln\)) as the new base. For this problem, we'll use base 10.

Step 2: Apply the Change of Base Formula

Apply the formula using base 10 for both the numerator and the denominator: \[ \log_5 10 = \frac{\log_{10} 10}{\log_{10} 5} \] which simplifies to: \[ \log_5 10 = \frac{1}{\log_{10} 5} \] because \(\log_{10} 10 = 1\).

Step 3: Calculate the Logarithm of the Denominator

Now, calculate \(\log_{10} 5\) using a calculator that provides decimal approximations: \[ \log_{10} 5 \approx 0.69897 \]

Step 4: Compute the Final Value

Plug the value from the previous step into the equation: \[ \log_5 10 = \frac{1}{0.69897} \approx 1.4307 \] Thus, the value of \(\log_5 10\) is approximately 1.4307.

Key Concepts

Logarithms

Logarithms are a way to express how many times a number, known as the base, must be multiplied by itself to reach another number. This concept is fundamental in mathematics and is especially useful in dealing with large numbers or when solving equations involving exponential growth or decay. In essence, if you have

• \( b^x = a \),
• then \( x = \log_b a \).

The logarithm tells us the exponent \( x \).
For instance, \( \log_{10} 100 \) equals 2 because 10 raised to the power of 2 gives 100. When you see "log," it usually implies base 10 if not specified otherwise. It's important to note that logarithms can have different bases, like 2, 5, or even \( e \) (approximately 2.718), which is used in natural logarithms. Understanding this concept is crucial for tasks such as changing the base of a logarithm, which can simplify calculations and reveal new properties.

Base Conversion

Changing the base of a logarithm can make computations more straightforward. The change of base formula is a powerful tool in logarithmic mathematics. It allows one to convert a challenging logarithm problem into a simpler one with a more familiar base. The formula is:

• \( \log_b a = \frac{\log_c a}{\log_c b} \),

where \( c \) is a more convenient base, usually 10 or \( e \).
In practice, this means you can take a logarithm that might be difficult to calculate directly, like \( \log_5 10 \), and use a calculator to find \( \log_{10} 10 \) and \( \log_{10} 5 \) instead. This method eliminates the need to perform complex base-specific calculations.
Why use base 10?
Electronic calculators commonly have pre-set functionality for base 10 (common logarithms) and base \( e \) (natural logarithms), making it easier and faster to perform base conversions and calculations. Thus, mastering base conversion through the change of base formula can significantly enhance one's ability to solve logarithmic problems efficiently.

Numerical Approximation

Numerical approximation is an essential technique, often used when the exact solution to a mathematical problem is challenging or impossible to obtain manually. In the context of logarithms, it involves using calculators or computer algorithms to find approximate solutions to logarithmic expressions. This is particularly useful when dealing with logarithms of numbers that do not result in neat whole numbers.

• For instance, \( \log_{10} 5 \) doesn't yield a simple integer.
• Instead, with computational tools, we approximate it to \( 0.69897 \).

Using approximations, one can solve real-world problems where precision to several decimal places is sufficient. Calculators today are so advanced that they can deliver high precision logarithmic values, such as four decimal places, that are acceptable for most practical purposes. This involves dividing precise values, as shown in changing \( \log_5 10 \) to a decimal by using its approximate base 10 logarithms, yielding an answer like \( 1.4307 \). Essentially, numerical approximation bridges the gap between algebraic representation and practical solution-finding.