Exponential Functions - Problem Solving

An exponential function is a function of the form \(f(x)=a \cdot b^x,\) where \(a\) and \(b\) are real numbers and \(b\) is positive. Exponential functions are used to model relationships with exponential growth or decay. Exponential growth occurs when a function's rate of change is proportional to the function's current value. Whenever an exponential function is decreasing, this is often referred to as exponential decay.

To solve problems on this page, you should be familiar with

• rules of exponents - algebraic
• solving exponential equations
• graphs of exponential functions.

Suppose that the population of rabbits increases by 1.5 times a month. When the initial population is 100, what is the approximate integer population after a year?

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The population after \(n\) months is given by \(100 \times 1.5^n.\) Therefore, the approximate population after a year is \[100 \times 1.5^{12} \approx 100 \times 129.75 = 12975. \ _\square \]

Suppose that the population of rabbits increases by 1.5 times a month. At the end of a month, 10 rabbits immigrate in. When the initial population is 100, what is the approximate integer population after a year?

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Let \(p(n)\) be the population after \(n\) months. Then \[p(n+2) = 1.5 p(n+1) + 10\] and \[p(n+1) = 1.5 p(n) + 10,\] from which we have \[p(n+2) - p(n+1) = 1.5 \big(p(n+1) - p(n)\big).\] Then the population after \(n\) months is given by \[p(0) + \big(p(1) - p(0)\big) \frac{1.5^{n} - 1}{1.5 - 1} .\] Therefore, the population after a year is given by \[\begin{align} 100 + (160 - 100) \frac{1.5^{12} - 1}{1.5 - 1} \approx& 100 + 60 \times 257.493 \\ \approx& 15550. \ _\square \end{align}\]

Suppose that the annual interest is 3 %. When the initial balance is 1,000 dollars, how many years would it take to have 10,000 dollars?

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The balance after \(n\) years is given by \(1000 \times 1.03^n.\) To have the balance 10,000 dollars, we need \[\begin{align} 1000 \times 1.03^n \ge& 10000 \\ 1.03^n \ge& 10\\ n \log_{10}{1.03} \ge& 1 \\ n \ge& 77.898\dots. \end{align}\] Therefore, it would take 78 years. \( _\square \)

The half-life of carbon-14 is approximately 5730 years. Humans began agriculture approximately ten thousand years ago. If we had 1 kg of carbon-14 at that moment, how much carbon-14 in grams would we have now?

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The weight of carbon-14 after \(n\) years is given by \(1000 \times \left( \frac{1}{2} \right)^{\frac{n}{5730}}\) in grams. Therefore, the weight after 10000 years is given by \[1000 \times \left( \frac{1}{2} \right)^{\frac{10000}{5730}} \approx 1000 \times 0.298 = 298.\] Therefore, we would have approximately 298 g. \( _\square \)