Problem 26 - Intermediate Algebra

Question

For Problems \(23-32\), approximate each of the following logarithms to three decimal places. $$ \log _{5} 1.4 $$

Step-by-Step Solution

Answer

\( \log_{5} 1.4 \approx 0.209 \).

Step 1: Use Change of Base Formula

We use the change of base formula to solve the problem since our calculator might not have a logarithm base 5 function. The change of base formula is:\[ \log_a b = \frac{\log_c b}{\log_c a} \]We will use base 10 (common logarithm) for simplicity. Thus, we have:\[ \log_5 1.4 = \frac{\log_{10} 1.4}{\log_{10} 5} \]

Step 2: Calculate the Numerator

Calculate \( \log_{10} 1.4 \) using a calculator. This will give us the logarithm of 1.4 in base 10.Upon calculation, we find that:\[ \log_{10} 1.4 \approx 0.146 \]

Step 3: Calculate the Denominator

Next, we calculate \( \log_{10} 5 \) using a calculator. This will give us the logarithm of 5 in base 10.Upon calculation, we find that:\[ \log_{10} 5 \approx 0.699 \]

Step 4: Divide Results to Find \( \log_5 1.4 \)

We now divide the result of the numerator by the result of the denominator:\[ \log_5 1.4 = \frac{0.146}{0.699} \approx 0.209 \]

Step 5: Approximate to Three Decimal Places

The result from the previous step is already rounded to three decimal places. Therefore, \( \log_{5} 1.4 \approx 0.209 \).

Key Concepts

Change of Base Formula

The change of base formula is a helpful tool when working with logarithms. At times, calculators do not directly support logarithms with bases other than 10 or 2. This formula allows us to convert a logarithm with any base into a division of logarithms using a base our calculator understands.
Using the formula looks like this:

• Change base formula: \( \log_a b = \frac{\log_c b}{\log_c a} \)
• Here, \( a \) is the base we are changing from, \( b \) is the number we are taking the log of, and \( c \) is the new base, often 10 or \( e \).

In our problem, we are converting a base 5 logarithm into a base 10 logarithm, because most calculators come equipped with base 10 options. By applying this, we find:\[ \log_5 1.4 = \frac{\log_{10} 1.4}{\log_{10} 5} \] This equivalence lets us easily compute the value using standard calculator functions. This flexibility to change bases simplifies calculations and expands the use of logarithms in various applications.

Common Logarithm

A common logarithm is a logarithm where the base is 10. It is widely used in science and engineering fields. When you see \( \log \) by itself, it usually implies \( \log_{10} \). Calculators frequently have a button for this type of logarithm in particular.
To understand better:

• \( \log_{10} x \) is the power to which 10 must be raised to equal \( x \).
• It is often simply written as \( \log x \).

In this exercise, we first calculate \( \log_{10} 1.4 \) and \( \log_{10} 5 \) as part of using the change of base formula. The approximation of \( \log_{10} 1.4 \) is approximately 0.146, while \( \log_{10} 5 \) comes to around 0.699. These calculations allow us to determine the exact value of \( \log_5 1.4 \) without directly needing a base 5 logarithm function.

Numerical Approximation

Numerical approximation is an important mathematical concept which involves estimating a number when it is not possible or necessary to compute the exact value. This is particularly useful when working with logarithms that involve non-integer bases or unusual numbers.
Such approximations are carried out to a specific number of decimal places to cater to practical applications, such as engineering or graphics.
For our example:

• We calculated \( \log_{10} 1.4 \approx 0.146 \) and \( \log_{10} 5 \approx 0.699 \).
• The division \( \frac{0.146}{0.699} \approx 0.209 \) results in the final answer.

The result was rounded to three decimal places, meeting the problem's specified level of precision. This precision helps provide clear and usable data for analysis and broader calculations in diverse fields.