Tutorial: Exponential functions and models

(This topic is also in Section 2.2 in Finite Mathematics and Applied Calculus)

#[I don't like this new tutorial. Take me back to the older tutorial!][No me gusta este nueve tutorial. ¡Regresame al tutorial más viejo (solo en inglés)!]#

Resources

Function evaluator and grapher

Excel grapher

Note To work effectively with exponential functions, we need to understand exponents and also know the laws of exponents. If you are not comfortable with these requirements, visit the %%exponentstut.

Warmup: Review of exponent laws

The following lists, taken from that %%exponentstut, gives the laws of exponents we will be using.
We use the following rules to combine exponential epxressions with the same base:

Exponent laws to combine expressions with the same base

#[Law][Ley]#

%%Examples

#[Comments][Comentarios]#

1. $a^ma^n = a^{m+n}$

$2^32^2 = 2^{3+2} = 2^5 = 32$
$2^32^{-2} = 2^{3-2} = 2^1 = 2$
$(-9)^4(-9)^{-2} = (-9)^{4-2} = (-9)^2 = 81$
$x^3x^4 = x^{3+4} = x^7$
$x^{-4}x^3 = x^{-4+3} = x^{-1} = \dfrac{1}{x}$

If the bases in a product match, add the exponents.
If the bases do not match the rule does not apply.
To see why this rule holds in the case of positive exponents, notice that

$a^na^m$	$= (a \cdot a\cdot a\ \cdots \ a)(a \cdot a\cdot a\ \cdots \ a)$
	$\qquad \quad n$ #[times][veces]# $\quad \ m$ #[times][veces]#
	$= a \cdot a\cdot a\ \cdots \ \cdots \ a$
	$\quad$ $n+m$ #[times][veces]#

#[The rule does not apply to sums: ][La regla no se aplica a sumas: ]#
$\qquad \quad a^m+a^n \neq (a+a)^n$

2. $\dfrac{a^m}{a^n} = a^{m-n}$
(%%if $a \neq 0$)

$\dfrac{2^3}{2^2} = 2^{3-2} = 2^1 = 2$
$\dfrac{x^3}{x^4} = x^{3-4} = x^{-1} = \dfrac{1}{x}$
$\dfrac{2^3}{2^{-2}} = 2^{3-(-2)} = 2^5 = 32$
$\dfrac{x^{-4}}{x^{-3}} = x^{-4-(-3)} = x^{-1} = \dfrac{1}{x}$
$\dfrac{1}{x^{-3}} = \dfrac{x^0}{x^{-3}} = x^{0-(-3)} = x^3$

If the bases in a quotient match, subtract the exponents.
If the bases do not match the rule does not apply.
Notice that this rule follows from cancellation in the case of positive exponents.
The rule does not apply to differences:
$\qquad a^m-a^n \neq a^{m-n}$

3. $\dfrac{1}{a^n} = a^{-n}$
(%%if $a \neq 0$)

$\dfrac{1}{2^3} = 2^{-3}$
$5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}$
$\dfrac{1}{5^{-2}} = 5^{-(-2)} = 5^2 = 25$

Rule 3 is actually a special case of Rule 2:

$\dfrac{1}{a^n}$	$=\dfrac{a^0}{a^n}$	#[Meaning of zero exponent][Significado del exponente cero]#
	$=a^{0-n}$	#[Rule 2][Regla 2]#
	$=a^{-n}$

4. $(a^n)^m = a^{nm}\ $

$(2^3)^2 = 2^{3\times 2} = 2^{6} = 64$
$(x^3)^4 = x^{3 \times 4} = x^{12}$
$(2^{-3})^2 = 2^{(-3)\times2} = 2^{-6} = \dfrac{1}{64}$
$(x^{-3})^{-4} = x^{(-3)\times(-4)} = x^{12}$

#[Raising a power to a power corresponds to mutliplying the powers.][Elevar una potencia a una potencia corresponde a multiplicar las potencias.]#

Suggested video for this topic: Video by Mometrix Academy

Now some for you to do:

Exponent laws to combine expressions with the different bases

#[Rule][Regla]#

%%Examples

#[Comments][Comentarios]#

1. $(ab)^n = a^nb^n$

$(2 \cdot 3)^2 = 2^2 \cdot 3^2 = 4 \times 9 = 36$
$(2 \cdot 3)^{-2} = 2^{-2} \cdot 3^{-2} = \dfrac{1}{4} \times \dfrac{1}{9} = \dfrac{1}{36}$
$(4(-3))^{2} = 4^2 \cdot (-3)^2 = 16 \times 9 = 144$
$(xy)^{-4} = x^{-4}y^{-4}$
$(-xy)^3 = (-x)^3(y)^3$

The $n$th power of a product is the product of the $n$th powers.
To see why this rule holds in the case of positive exponents, notice that

$(ab)^n$=	$(ab \cdot ab \cdot ab \ \cdots \ ab )$
	$\qquad \quad n$ #[times][veces]#
=	$(a \cdot a\cdot a\ \cdots \ a)(b \cdot b\cdot b\ \cdots \ b)$
=	$a^nb^n$

#[The rule does not apply to sums: ][La regla no se aplica a sumas: ]#
$\qquad (a+b)^n \neq a^n + b^n$

2. $\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}$
(%if $b \neq 0$)

$\left(\dfrac{3}{2}\right)^4 = \dfrac{3^4}{2^4} = \dfrac{81}{16}$
$\left(\dfrac{x}{y}\right)^{-2} = \dfrac{x^{-2}}{y^{-2}}$
$\left(\dfrac{1}{y}\right)^3 = \dfrac{1^3}{y^3} = \dfrac{1}{y^3}$
$\left(\dfrac{-2}{-3}\right)^2 = \dfrac{(-2)^2}{(-3)^2} = \dfrac{4}{9}$

The $n$th power of a quotient is the quotient of the $n$th powers. To see why this rule holds in the case of positive exponents, notice that

$\left(\dfrac{a}{b}\right)^n$=	$\left(\dfrac{a}{b} \cdot \dfrac{a}{b} \cdot \dfrac{a}{b} \ \cdots \ \dfrac{a}{b} \right)$
	$\qquad \quad n$ #[times][veces]#
=	$\dfrac{a \cdot a\cdot a\ \cdots \ a}{b \cdot b\cdot b\ \cdots \ b}$
=	$\dfrac{a^n}{b^n}$

The rule does not apply to differences:
$\qquad (a-b)^n \neq a^n - b^n$

Now some for you to do:

Exponential functions

Exponential functions are frequently used to model the growth or depreciation of financial investments, population growth, radioactive decay, and phenomena where a quantity is allowed to undergo unrestrained growth.

Exponential function

An exponential function has the form

$f(x) = Ab^x$ \gap[40] \t Function form \\ $y= Ab^x$ \gap[40] \t Equation form

where $A$ and $b$ are constants with $b$ positive. We call $b$ the base of the exponential function.

Technology formula: A*b^x

Suggested video for this topic: Video by ThinkwellVids
#[Note][Nota]# Online videos such as this one cover only the special case $f(x) = b^x$; that is, they take $A$ to be $1$.

Examples

\t $f(x) = 2^x$ \t $A = 1, b = 2$ \gap[20] \t Technology: 2^x \\ \t $f(3) = 2^3 = 8$ \\ \t $f(0) = 2^0 = 1$ \\ \t $\displaystyle f(-4) = 2^{-4} = \frac{1}{2^4} = \frac{1}{16}$ \\ \\

\t $g(x) = 10(1.5^x)$ \t $A = 10, b = 1.5$ \gap[20] \t Technology: 10*1.5^x \\ \t !2! $g(2) = 10(1.5^2) = 10(2.25) = 22.5$ \\ \t $\displaystyle g(-1) = 10(1.5^{-1}) = \frac{10}{1.5} = \frac{20}{3}$ \\ \\

\t $h(x) = 3(4^{-2x})$ \t Rewrite this as $3(4^{-2})^x = 3\left(\dfrac{1}{16}\right)^x$ \gap[20] \t Technology: 3*4^(-2x) \\ \t \t $A = 3, b = \dfrac{1}{16}$ \\

Exponential functions: Numerical and graphical viewpoints

Here is a table of values for the exonential function together with its graph $f(x) = 5(2^x); \quad (A = 5, b = 2)$:

Role of A and b
Notice that the $y$-intercept is $A = 5$ (obtained by setting $x = 0$). In general:

In the graph of $f(x) = Ab^x,$ $A$ is the $y$-intercept, or the value of $y$ when $x = 0$.

As for $b$, notice that the value of $y$ is multiplied by $b = 2$ for every increase of 1 unit in $x$:

The value of $y$ is multiplied by $b$ for every 1-unit increase of $x$:

$\qquad$ #[Multiply by 2][Multiplica por 2]#.

Distinguishing exponential change from linear change

We saw above that, in a tabular representation of an exponential fuction $f(x) = Ab^x$, the value of $f(x)$ is multiplied by $b$ for every increase of 1 unit in $x$. For instance, we looked at the following table for $f(x) = 5(2^x)$:

$\qquad$ #[Multiply by 2][Multiplica por 2]#.

#[By contrast, in a tabluar representation of a linear function $f(x) = mx + b$, the value of $f(x)$ is incremented by $m$ for every increase of 1 unit in $x$. For instance, a table of values for $g(x) = 2x + 5$ would look like this:][Por el contrario, en una representación tabular de una función lineal $f(x) = mx + b$, el valor de $f(x)$ se incrementa en $m$ por cada aumento de 1 unidad en $x$. Por ejemplo, una tabla de valores para $g(x) = 2x + 5 $ se ver�a as�]#

$\qquad$ #[Add 2][Suma 2]#.

Finding exponential functions from data

To model real life situations with exponential functions, as with the case of linear functions, all we need are two data points. The following procedure shows how to find the needed equation.

The exponential curve through two points: An example

Let's find an equation of the exponential curve $y = Ab^x$ through $(1, 10.4)$ and $(3, 41.6)$.
Step 1 Substitute the coordinates of the given points in the equation $y = Ab^x$:

$10.4 = Ab^1$ \t Substitute $(1, 10.4)$. \\ $41.6 = Ab^3.\qquad$ \t Substitute $(3, 41.6)$.

Step 2 Divide the second equation by the first to eliminate the constant $A$ ad solve for $b$:

$\displaystyle \frac{41.6}{10.4} = \frac{Ab^3}{Ab}$ = $b^2$ \\ $\displaystyle b^2 = \frac{41.6}{10.4} = 4$ \\ $\displaystyle b = 4^{1/2} = 2$ \t Take reciprocal power of both sides.

Step 3 Substitute for $b$ in either equation to obtain $A$:

$10.4 = A(2^1) = 2A \qquad$ \t Substitute $b = 2$ in the first equation above: $10.4 = Ab^1$ \\ $\displaystyle A = \frac{10.4}{2} = 5.2$

#[Thus we have $A = 5,2$ and $b = 2$, so that our desired equation is][Por lo tanto, tenemos $A = 5.2$ y $b = 2$, de modo que nuestra ecuación deseada es]#

$y = 5.2(2^x). \qquad$ \t $y = Ab^x$

Suggested video for this topic: Video by Phil Clark

Applications: Exponential models

We mentioned at the start of this tutorial that exponential functions are used to model growth or depreciation of financial investments, population growth and radioactive decay. We will now see how this is done.

Exponential growth and decay models

A quantity $y=Ab^t$ varying with exponentially with time $t$ experiences exponential growth if $b \gt 1$ and exponential decay if $b \lt 1$.

$y = Ae^{bt}; \ \ b \gt 1$
#[Exponential growth][Crecimiento exponencial]#

$y = Ae^{bt}; \ \ b \lt 1$
#[Exponential decay][Decaimiento exponencial]#

%%Examples

1. #[Compound interest][Interés compuesto]# (#[See the][Ve el]# %%modelstut.)
If an amount (present value) $P$ is invested for $t$ years at an annual rate of $r$, and if the interest is compounded (reinvested) $m$ times per year, then the future value $F$ is

$\displaystyle F=P\left(1+\frac{r}{m}\right)^{mt}$. For instance, if &D&2,000 is invested for $t$ years at a 6% rate of interest, and the interest is reinvested every month (so $m = 12$), then

$\displaystyle F=2{,}000\left(1+\frac{0.06}{12}\right)^{12t}=2{,}000(1.005)^{12t}$, which we can rewrite as an exponential function of $t$:

$A = 2{,}000,\ b = 1.005^{12}.$

As $b$ is greater than 1, this is an exponential growth model.

2. Radioactive decay and carbon dating (#[Also see the][Ve también el]# %%modelstut.)

Carbon 14, an unstable isotope of carbon, decays extremely slowly to nitrogen and is used in the dating of fossils. The amount of carbon 14 remaining in a sample that originally contained $A$ grams is given by an exponential function of time $t$:

$C(t) = A(0.999879)^t \qquad$ Exponential decay; $b = 0.999879$

#[where $t$ is time in years.][donde $t$ es tiempo en años.]#

1. After 10,000 years, a sample that originally contained 100g of carbon 14 will still contain

$C(10{,}000) = 100(0.999879)^{10{,}000} \approx 29.8$ g #[carbon][carbono]# 14.

2. If, after 10,000 years, the amount of carbon 14 in a sample is found to be 50 g, how much did it originally contain?

To find the original amount, substitute the given information in the formula:

$50 = A(0.999879)^{10{,}000}$ \\ $A = \dfrac{50}{0.999879^{10{,}000}} \approx 167.7$ g

3. Dating a sample: If a sample originally containing 100g of carbon 14 has decayed to half that amount, how old is the sample?

#[Again, substitute the given information in the formula:][Otra vez, sustituya la información dada en la fórmula:]#

$50 = 100(0.999879)^{t}$ \\ $\dfrac{1}{2} = 0.999879^t$ \t #[Divide both sides by 100.][Divide ambos lados por 100.]# \\ $t = \log_{0.999879}\left(\frac{1}{2}\right)\approx 5728$ years \gap[40] \t #[Rewrite in logarithmic form.][Reescribir en forma logarítmica.]#

Now try some of the exercises in Section 2.2 in Finite Mathematics and Applied Calculus. or move ahead by pressing "Next tutorial" on the sidebar.