In mathematical modeling, we choose a familiar general function with properties that suggest that it will model the real-world phenomenon we wish to analyze. In the case of rapid growth, we may choose the exponential growth function:

y=A_{0}e^{kt}

where A_{0} is equal to the value at time zero (the starting point), e is Euler’s constant, and k is a positive constant that determines the rate (percentage) of growth. We may use the exponential growth function in applications involving doubling time, the time it takes for a quantity to double. Such phenomena as wildlife populations, financial investments, biological samples, and natural resources may exhibit growth based on a doubling time.

Characteristics of exponential function:

An exponential function with the form y=A_{0}e^{kt} has the following characteristics:

one-to-one function
horizontal asymptote: y=0
domain: (-\infty,\, \infty)
range: (0,\,\infty)
x-intercept: none
y-intercept: (0,\,A_{0})
increasing if k \gt 0
decreasing if k \lt 0

How to: Half-life

Given the half-life, find the decay rate:

Write A=A_{0}e^{kt}
Replace A by \dfrac{1}{2}A_{0} and replace t by the given half-life.
Solve to find k. Express k as an exact value

Note: It is also possible to find the decay rate using k=\dfrac{\ln \frac{1}{2}}{t}

Examples

Example 1

A population of bacteria doubles every hour. If the culture started with 10 bacteria, what is the constant that determines the rate (percentage) of growth.

Worked Solution

Create a strategy

We can use the exponential growth function: y=A_{0}e^{kt}, where A_{0} is equal to the value at time zero (the starting point), e is Euler’s constant, and k is a positive constant that determines the rate (percentage) of growth.

Apply the idea

To find the k value, we must substitute the known values into the exponential equation. To find A_{0} we use the fact that A_{0} is the amount at the starting point; therefore, A_{0}=10. We know what happens after one hour; the culture doubles. So, the t=1 hour and y=20.

\displaystyle y	\displaystyle =	\displaystyle A_{0}e^{kt}	Write the exponential growth function
\displaystyle 20	\displaystyle =	\displaystyle (10)e^{k \times 1}	Substitute y=20,\,A_{0}=10 and t=1
\displaystyle 2	\displaystyle =	\displaystyle e^{k}	Divide both sides by 10
\displaystyle \ln {2}	\displaystyle =	\displaystyle k	Take the natural logarithm

Example 2

The half-life of carbon - 14 is 5730 years. Express the amount of carbon - 14 remaining as a function of time, t.

Worked Solution

Create a strategy

To find the half-life of a function describing exponential decay, we can use the following equation: \dfrac{1}{2}A_{0}=A_{0}e^{kt}.

Apply the idea

\displaystyle \dfrac{1}{2}A_{0}	\displaystyle =	\displaystyle A_{0}e^{kt}	Write the exponential decay equation
\displaystyle \dfrac{1}{2}A_{0}	\displaystyle =	\displaystyle A_{0}e^{k \times 5730}	Substitute t=5730
\displaystyle \dfrac{1}{2}	\displaystyle =	\displaystyle e^{5730 k}	Divide both sides by A_{0}
\displaystyle \ln \left(\dfrac{1}{2}\right)	\displaystyle =	\displaystyle 5730 k	Take the natural log
\displaystyle k	\displaystyle =	\displaystyle \dfrac{\ln\left(\frac{1}{2}\right)}{5730}	Divide both sides by 5730

To express the remaining as a function of time, rewrite the formula replacing the k with value above:

\displaystyle A	\displaystyle =	\displaystyle \dfrac{1}{2} A_{0}
\displaystyle A	\displaystyle =	\displaystyle A_{0}e^{\frac{\ln \frac{1}{2}}{5730}}

Reflect and check

When we are measuring decay, it is expected that the k value be negative. Lets find the approximate value of the k using the calculator: k is approximately -1.209\,7 \times 10^{-4}.

Example 3

How long will it take for 10\% of a 1000-gram sample of uranium-235 to decay? Uranium-235's half-life is 703\,800\,000 years.

Worked Solution

Create a strategy

We can use the exponential decay function: A=A_{0}e^{kt} and k=\dfrac{\ln \frac{1}{2}}{t}.

Apply the idea

We are looking for the time it will take for 10\% of the substance to decay. Therefore, the starting point A_{0} = 1000 and the ending point is A=0.9 \times 1000 which is A=900.

\displaystyle A	\displaystyle =	\displaystyle A_{0}e^{kt}	Write the exponential decay function
\displaystyle 900	\displaystyle =	\displaystyle 1000 e^{kt}	Substitute A=900 and A_{0}=1000
\displaystyle k	\displaystyle =	\displaystyle \dfrac{\ln \left(\frac{1}{2}\right)}{t}	Write the exponential decay function
\displaystyle k	\displaystyle =	\displaystyle \dfrac{\ln \left(\frac{1}{2}\right)}{703\,800\,000}	Substitute t=703\,800\,000
\displaystyle 900	\displaystyle =	\displaystyle 1000 e^{\frac{\ln \left(\frac{1}{2}\right)}{703\,800\,000}\times t}	Substitute k=\dfrac{\ln \left(\frac{1}{2}\right)}{703\,800\,000}
\displaystyle 0.9	\displaystyle =	\displaystyle e^{\frac{\ln \left(\frac{1}{2}\right)}{703\,800\,000}\times t}	Divide both sides by 1000
\displaystyle \ln (0.9)	\displaystyle =	\displaystyle {\frac{\ln \left(\frac{1}{2}\right)}{703\,800\,000}\times t}	Take the natural log
\displaystyle t	\displaystyle =	\displaystyle \dfrac{(703\,800\,000)\times \ln(0.9)}{\ln \left(\frac{1}{2}\right)}	Multiply both sides by \dfrac{703\,800\,000}{\ln \left(\frac{1}{2}\right)}
\displaystyle t	\displaystyle \approx	\displaystyle 106\,979\,777	Simplify

Idea summary

Characteristics of exponential function:

An exponential function with the form y=A_{0}e^{kt} has the following characteristics:

one-to-one function
horizontal asymptote: y=0
domain: (-\infty,\, \infty)
range: (0,\,\infty)
x-intercept: none
y-intercept: (0,\,A_{0})
increasing if k \gt 0
decreasing if k \lt 0

How to: Half-life

Given the half-life, find the decay rate:

Write A=A_{0}e^{kt}
Replace A by \frac{1}{2}A_{0} and replace t by the given half-life.
Solve to find k. Express k as an exact value

Note: It is also possible to find the decay rate using k=\dfrac{\ln \frac{1}{2}}{t}

Appropriate Model for Data

Now that we have discussed various mathematical models, we need to learn how to choose the appropriate model for the raw data we have. Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself. Not all data can be described by elementary functions. Sometimes, a function is chosen that approximates the data over a given interval. For instance, suppose data were gathered on the number of homes bought in the United States from the years 1960 to 2013. After plotting these data in a scatter plot, we notice that the shape of the data from the years 2000 to 2013 follow a logarithmic curve. We could restrict the interval from 2000 to 2010, apply regression analysis using a logarithmic model, and use it to predict the number of home buyers for the year 2015.

Examples

Example 4

pH is a measure of how acidic or alkaline a substance is, and the pH scale goes from 0 to 14,\,0 being most acidic and 14 being most alkaline. Water in a stream has a neutral pH of about 7. The pH (p) of a substance can be found according to the formula p=-\log_{10} h, where h is the substance’s hydrogen ion concentration.

Store-bought apple juice has a hydrogen ion concentration of about h=0.000\,2. Determine the pH of the apple juice correct to one decimal place.

Worked Solution

Create a strategy

Substitute the value of h=0.000\,2 in the formula.

Apply the idea

\displaystyle p	\displaystyle =	\displaystyle -\log_{10}h	Write the formula
	\displaystyle =	\displaystyle -\log_{10}{0.000\,2}	Substitute h=0.000\,2
	\displaystyle =	\displaystyle 3.7	Evaluate

Is the apple juice acidic or alkaline?

Worked Solution

Create a strategy

Remember that if the pH of apple juice is between 0 and 7 (not inclusive), then apple juice is acidic. If the pH of apple juice is between 7 (not inclusive) and 14, then apple juice is alkaline.

Apply the idea

The pH of the apple juice is 3.7.

So, the apple juice is acidic.

A banana has a pH of about 8.3. Solve for h, its hydrogen ion concentration, leaving your answer as an exact value.

Worked Solution

Create a strategy

Use the formula: p=-\log_{10}{h}.

Apply the idea

\displaystyle p	\displaystyle =	\displaystyle -\log_{10}{h}	Write the formula
\displaystyle 8.3	\displaystyle =	\displaystyle -\log_{10}{h}	Substitute p=8.3
\displaystyle \log_{10}{h}	\displaystyle =	\displaystyle -8.3	Multiply both sides by -1
\displaystyle h	\displaystyle =	\displaystyle 10^{-8.3}	Rewrite as b=a^{n}

Idea summary

Many factors influence the choice of a mathematical model, among which are experience, scientific laws, and patterns in the data itself.

2.14 Logarithmic function context and data modeling

Introduction

Ideas

Practical problems with exponential and logarithms

Examples

Example 1

Example 2

Example 3

Appropriate Model for Data

Examples

Example 4

Outcomes

2.14.A

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