Problem 110 - Elementary and Intermediate Algebra

Question

The formula \(t=\frac{\ln 2}{r}\) gives the time \(t\) for a population to double, where \(r\) is the annual rate of continuous compounding. Write the formula in an equivalent form so that it involves a common logarithm, not a natural logarithm.

Step-by-Step Solution

Answer

Equivalent formula: \(t = \frac{\log_{10} 2}{r \cdot \log_{10} e}\).

Step 1: Understand the Conversion Between Logarithms

The given formula uses the natural logarithm, denoted by \(\ln\), which is the logarithm to base \(e\). The problem asks to express it using common logarithms, denoted by \(\log\), which is the logarithm to base 10. To convert \(\ln\) to \(\log\), we use the change of base formula: \(\ln x = \log_e x = \frac{\log_{10} x}{\log_{10} e}\).

Step 2: Apply the Change of Base Formula

Substitute \(\ln 2\) with \(\frac{\log_{10} 2}{\log_{10} e}\) in the original formula. Thus, the formula becomes: \[t = \frac{\frac{\log_{10} 2}{\log_{10} e}}{r}\]

Step 3: Simplify the Expression

Rewrite the expression from Step 2 by rearranging the division as multiplication by the reciprocal. Thus, our formula becomes:\[t = \frac{\log_{10} 2}{r \cdot \log_{10} e}\]This form involves the common logarithm \(\log_{10}\).

Key Concepts

Exponential Growth

Exponential growth is a vital concept often encountered in real-life contexts such as population growth, radioactive decay, and investment growth over time. When something grows exponentially, it means that the rate of growth is proportional to the current value, resulting in the quantity increasing over time. This type of growth is characterized by a constant multiplier over equal time intervals.

In exponential growth, a key feature is the doubling time, which is the time it takes for a quantity to double in size. This doubling time can be calculated using logarithms, particularly when dealing with continuous compounding. The formula \(t = \frac{\ln 2}{r}\) describes the doubling time \(t\) when the growth rate \(r\) is given.

This specific formula uses a natural logarithm because natural logarithms simplify the calculations involved in continuous growth processes. However, understanding how to switch between natural and common logarithms is helpful, as it allows a clearer interpretation with base 10, which many find more intuitive.

Natural Logarithm

The natural logarithm, represented by \(\ln\), is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. This logarithm is particularly useful in mathematics when we deal with growth processes that are continuous, such as population growth or interest compounding.

Natural logarithms help in simplifying equations involving exponential increase or decay. For instance, when we are dealing with continuous compounding interest, \(\ln\) provides a neat and compact form for the expressions. In our original formula \(t = \frac{\ln 2}{r}\), the \(\ln\) helps to directly relate growth rates and doubling time in a straightforward manner.

It’s important to remember that the natural logarithm can be converted into a common logarithm using the change of base formula: \(\ln x = \frac{\log_{10} x}{\log_{10} e}\). This formula is often used to switch between logarithmic bases, allowing more flexible and intuitive calculations.

Common Logarithm

The common logarithm, denoted by \(\log_{10}\), is a logarithm with base 10. It is widely used because the base of 10 is relatable to decimal numbers, making it easier to comprehend for practical applications. This type is especially prominent in fields like chemistry and engineering, where base 10 logarithms simplify measurements and calculations.

In the context of the exercise where the natural logarithm needs to be converted to a common logarithm, we utilize the change of base formula. By transforming the natural logarithm \(\ln 2\) into \(\frac{\log_{10} 2}{\log_{10} e}\), the formula becomes easier to interpret and apply in contexts where base 10 is more appropriate.

Using common logarithms in the formula \(t = \frac{\log_{10} 2}{r \cdot \log_{10} e}\) not only changes the base but also aids in simplifying calculations and facilitating more understandable results for those who are more familiar with base 10 operations. This conversion is crucial for creating interchangeable expressions that can adapt to user preference or situational demands.