# 4.7 Exponential and logarithmic models  (Page 6/16)

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Using the model in [link] , estimate the number of cases of flu on day 15.

895 cases on day 15

## Choosing an 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.

Three kinds of functions that are often useful in mathematical models are linear functions, exponential functions, and logarithmic functions. If the data lies on a straight line, or seems to lie approximately along a straight line, a linear model may be best. If the data is non-linear, we often consider an exponential or logarithmic model, though other models, such as quadratic models, may also be considered.

In choosing between an exponential model and a logarithmic model, we look at the way the data curves. This is called the concavity. If we draw a line between two data points, and all (or most) of the data between those two points lies above that line, we say the curve is concave down. We can think of it as a bowl that bends downward and therefore cannot hold water. If all (or most) of the data between those two points lies below the line, we say the curve is concave up. In this case, we can think of a bowl that bends upward and can therefore hold water. An exponential curve, whether rising or falling, whether representing growth or decay, is always concave up away from its horizontal asymptote. A logarithmic curve is always concave away from its vertical asymptote. In the case of positive data, which is the most common case, an exponential curve is always concave up, and a logarithmic curve always concave down.

A logistic curve changes concavity. It starts out concave up and then changes to concave down beyond a certain point, called a point of inflection.

After using the graph to help us choose a type of function to use as a model, we substitute points, and solve to find the parameters. We reduce round-off error by choosing points as far apart as possible.

## Choosing a mathematical model

Does a linear, exponential, logarithmic, or logistic model best fit the values listed in [link] ? Find the model, and use a graph to check your choice.

 $x$ 1 2 3 4 5 6 7 8 9 $y$ 0 1.386 2.197 2.773 3.219 3.584 3.892 4.159 4.394

First, plot the data on a graph as in [link] . For the purpose of graphing, round the data to two significant digits.

Clearly, the points do not lie on a straight line, so we reject a linear model. If we draw a line between any two of the points, most or all of the points between those two points lie above the line, so the graph is concave down, suggesting a logarithmic model. We can try $\text{\hspace{0.17em}}y=a\mathrm{ln}\left(bx\right).\text{\hspace{0.17em}}$ Plugging in the first point, $\text{\hspace{0.17em}}\left(\text{1,0}\right)\text{,}\text{\hspace{0.17em}}$ gives $\text{\hspace{0.17em}}0=a\mathrm{ln}b.\text{\hspace{0.17em}}$ We reject the case that $\text{\hspace{0.17em}}a=0\text{\hspace{0.17em}}$ (if it were, all outputs would be 0), so we know $\text{\hspace{0.17em}}\mathrm{ln}\left(b\right)=0.\text{\hspace{0.17em}}$ Thus $\text{\hspace{0.17em}}b=1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=a\mathrm{ln}\left(\text{x}\right).\text{\hspace{0.17em}}$ Next we can use the point $\text{\hspace{0.17em}}\left(\text{9,4}\text{.394}\right)\text{\hspace{0.17em}}$ to solve for $\text{\hspace{0.17em}}a:$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=a\mathrm{ln}\left(x\right)\hfill \\ 4.394=a\mathrm{ln}\left(9\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}a=\frac{4.394}{\mathrm{ln}\left(9\right)}\hfill \end{array}$

Because $\text{\hspace{0.17em}}a=\frac{4.394}{\mathrm{ln}\left(9\right)}\approx 2,$ an appropriate model for the data is $\text{\hspace{0.17em}}y=2\mathrm{ln}\left(x\right).$

To check the accuracy of the model, we graph the function together with the given points as in [link] .

We can conclude that the model is a good fit to the data.

Compare [link] to the graph of $\text{\hspace{0.17em}}y=\mathrm{ln}\left({x}^{2}\right)\text{\hspace{0.17em}}$ shown in [link] .

The graphs appear to be identical when $\text{\hspace{0.17em}}x>0.\text{\hspace{0.17em}}$ A quick check confirms this conclusion: $\text{\hspace{0.17em}}y=\mathrm{ln}\left({x}^{2}\right)=2\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x>0.$

However, if $\text{\hspace{0.17em}}x<0,$ the graph of $\text{\hspace{0.17em}}y=\mathrm{ln}\left({x}^{2}\right)\text{\hspace{0.17em}}$ includes a “extra” branch, as shown in [link] . This occurs because, while $\text{\hspace{0.17em}}y=2\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ cannot have negative values in the domain (as such values would force the argument to be negative), the function $\text{\hspace{0.17em}}y=\mathrm{ln}\left({x}^{2}\right)\text{\hspace{0.17em}}$ can have negative domain values.

The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris