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Doubling time is a concept used to understand how quickly a quantity grows under specific growth conditions. It represents the period during which a given quantity doubles in size. In our problem, the number of Australian homes with computers doubles every 8 months. This means if you start with 100 homes, in 8 months you'll have 200. Doubling time provides a simple way to grasp how rapid growth is. It is a key way to measure exponential growth, where each new amount is a power of two. To estimate how fast something is growing exponentially, knowing the doubling time can be very insightful. In practice, you can use the formula:

• Doubling Time Formula: \( N = N_0 \times 2^{t/T} \)

Where:

• \( N \) is the final amount,
• \( N_0 \) is the initial amount,
• \( t \) is time, and
• \( T \) is the doubling time.

Natural logarithms are a key mathematical tool used to solve problems involving exponential growth or decay. In essence, logs are the inverse operations of exponentiation. For our purposes, the natural logarithm, denoted as \( \ln \), helps convert multiplication into addition, which simplifies solving exponential equations.When dealing with continuous growth, the equation \( 2 = e^{8k} \) comes up, named after the natural log base \( e \approx 2.718 \.\) In this problem, converting the doubling equation to determine the growth rate involves taking the natural logarithm on both sides:

• \( \ln(2) = 8k \).

Here, \( \ln(2) \approx 0.6931 \), giving a simple way to isolate the growth rate \( k \). The usage of \( \ln \) is vital when solving for variables in exponentials because it reshapes the problem to a linear form that is more easily handled. If you're facing exponential growth tasks, understanding natural logarithms is essential.

Continous growth rate is a concept often used to describe the rate at which a quantity increases over time in a smooth, unbroken manner. In our exponential growth example, the growth rate is consistent and continuous, not influenced by discrete jumps.Outlined in the mathematical model as \( N = N_0 e^{kt} \), where \( k \) is the continuous growth rate constant, represents how fast the initial value \( N_0 \) grows in a specified time \( t \). The task is finding this rate that ensures a twofold increase in one doubling period (8 months).The steps involved:

• Set \( N = 2N_0 \) for doubling.
• Substitute in \( 2 = e^{8k} \).

This setup lets us determine what \( k \) must be. Solving using natural logs, we obtained \( k = \frac{\ln(2)}{8} \), leading to a continuous growth rate of 0.0866 or 8.66% when expressed per month.

To make sense of growth in simpler terms, the percentage increase transforms the calculated continuous growth rate into a more relatable measure. It quantifies how much the initial amount grows per time unit in terms of percentage.Expressing growth rates as percentages is common in real-world applications, helping people understand the magnitude of growth easily. In this exercise, once we calculated \( k \approx 0.0866 \), converting it to a percentage involves a simple multiplication by 100, yielding the monthly percentage increase:

• \( 0.0866 \times 100 = 8.66\% \).

Now, this value, 8.66%, indicates how much the population of computer owners grows monthly. Turning abstract growth rates from calculations into percentages bridges a gap, making it intuitive and meaningful, whether referring to financial interest rates or population growth.