Problem 74 - Precalculus

Question

Determine whether each statement is true or false. $$\frac{\ln a}{\ln b}=\frac{\log a}{\log b}$$

Step-by-Step Solution

Answer

The statement is true.

Step 1: Understanding the Problem

We are given an expression \( \frac{\ln a}{\ln b} = \frac{\log a}{\log b} \) and need to determine whether it is true or false. The expression involves natural logarithms denoted by \( \ln \) and common (base 10) logarithms denoted by \( \log \).

Step 2: Introducing the Change of Base Formula

The change of base formula helps to convert logarithms from one base to another. It states that for any positive numbers \( a \) and \( b \): \[ \log_{c}(a) = \frac{\log_{k}(a)}{\log_{k}(c)} \] where \( k \) can be any valid base for logarithms.

Step 3: Applying the Change of Base Formula

Using the change of base formula, we can express the common logarithm as: \[ \log a = \frac{\ln a}{\ln 10} \] and \[ \log b = \frac{\ln b}{\ln 10} \] since \( \ln \) refers to the natural logarithm with base \( e \).

Step 4: Simplifying the Given Expression

Given the formulas obtained from the change of base, we substitute them into the original expression: \[ \frac{\log a}{\log b} = \frac{\frac{\ln a}{\ln 10}}{\frac{\ln b}{\ln 10}} = \frac{\ln a}{\ln b} \times \frac{\ln 10}{\ln 10} = \frac{\ln a}{\ln b} \] This shows the equality in the original expression, confirming that both sides are equal.

Step 5: Conclusion

Therefore, the expression \( \frac{\ln a}{\ln b} = \frac{\log a}{\log b} \) is true as verified by the change of base formula and our simplification.

Key Concepts

Change of Base Formula

The change of base formula is a powerful tool in logarithms that helps us switch from one base to another seamlessly. By converting logs into a base we are more comfortable with, we can solve problems more easily. This formula essentially states that for any positive numbers \( a \), \( b \), and a new base \( k \):

• \( \log_{c}(a) = \frac{\log_{k}(a)}{\log_{k}(c)} \) where \( k \) can be any base, like \( 10 \) or \( e \).

Using this formula, we can express logarithms from one base in terms of another. For example, converting a natural logarithm \( \ln a \) to a common logarithm \( \log a \) using base \( 10 \) is straightforward:

• \( \log a = \frac{\ln a}{\ln 10} \)

This transformation makes it possible to easily compare different types of logarithms or perform calculations where changing the base provides convenience.
For the expression \( \frac{\ln a}{\ln b} \) is equal to \( \frac{\log a}{\log b} \), using the change of base formulas for each, we see they equate, confirming the truth of the initial statement.

Natural Logarithm

The natural logarithm, usually denoted as \( \ln \), is a logarithm with base \( e \), where \( e \) is an irrational constant approximately equal to 2.718. The natural logarithm is especially useful in mathematical modeling and appears frequently in calculations involving growth and decay, among other applications.

• \( \ln(e) = 1 \) because any logarithm of its base is 1.
• The natural log of 1 is 0, since \( e^0 = 1 \).

It has unique properties, such as the fact that the derivative of \( \ln x \) is \( \frac{1}{x} \). These properties make it invaluable in calculus and other areas of higher mathematics.
In the context of our expression, \( \ln a \) and \( \ln b \) are used in the numerator and denominator respectively, and analyzing them using the change of base formula provided an easy way to relate \( \ln \) with \( \log \).

Common Logarithm

Common logarithms, known as \( \log \), utilize base \( 10 \). It is often used when dealing with decimal systems or wherever base 10 is more intuitive, such as in scientific notation or certain financial formulas.

• \( \log(10) = 1 \), because \( 10^1 = 10 \).
• The log of 1 is 0, since \( 10^0 = 1 \).

The common logarithm simplifies the process of dealing with powers of 10, which is why it's a staple in various scientific fields.
The conversion from a natural logarithm to a common logarithm, facilitated by the change of base formula, highlights its flexibility and interoperability. In the expression \( \frac{\ln a}{\ln b} = \frac{\log a}{\log b} \), the equality holds true thanks to the properties shared by all logarithms, regardless of their base.