Chapter 4.3 Symmetric Change

Exponential Functions

The function $f(x)=2x$, which is a linear function, models linear growth. For example, the difference between inputs that are a distance of $1$ apart is always $2$. That is \[f(n+1)-f(n)=2(n+1)-2n=2.\] In this specific example, it grows by the slope.

However, the function $g(x)=2^x$ is not linear. What is the difference in the function values between inputs that are a distance of $1$ apart? Here are some values of $g$:

$x$	$g(x)$
$0$	$1$
$1$	$2$
$2$	$4$
$3$	$8$
$4$	$16$

The differences between inputs that are a distance of $1$ apart are

\[g(1)-g(0)=1\]
\[g(2)-g(1)=2\]
\[g(3)-g(2)=4\]
\[g(4)-g(3)=8.\]

In this case, \[g(n+1)-g(n)=2^{n+1}-2^{n}=2^{n}.\]

This function expires what is called exponential growth.

Exponential Function

Take $b$ to be a number so that $b$ in $(0,1)\cup(1,\infty).$ The function \[f(x)=b^x\] is the exponential function with base $b$.

The domain of the exponential function is $\mathbb{R}$ and the range is $(0,\infty).$

We sometimes use the notation $\exp_b$ to denote it:

\[\exp_b(x)=b^x.\]

The exponential function is defined for all real numbers, but its range is not all real numbers. This is because there is no solution to the equation and the inequality \[b^x=0\quad\text{and}\quad b^x<0.\]

The one thing to keep in mind is that the exponential function and power function are not the same thing:

\[\exp_b(x)=b^x\quad\text{and}\quad{\rm{pow}}_n(x)=x^n.\]

The exponential function has a fixed base $b$, but the exponent or power varies. However, the power function has fixed exponent or power, but the base varies.

Understand this distinction with the following example.

Example 1

Compute the following.

$3^4$
$\left(\frac{1}{4}\right)^{2}$
$\left(\frac{1}{7}\right)^{-2}$
$\left(81\right)^{\frac{1}{2}}$

\[3^4=\underbrace{3\cdot 3\cdot 3\cdot}_{\text{multiply $3$ by itself $4$ times}}=81.\]
\[\left(\frac{1}{4}\right)^2=\underbrace{\frac{1}{4}\cdot\frac{1}{4}}_{\text{multiply$\frac{1}{4}$ by itself $2$ times}}=\frac{1}{16}.\]
\[\left(\frac{1}{7}\right)^{-2}=7^2=49.\]
\[\left(81\right)^{\frac{1}{2}}=\sqrt{81}=9.\]

Here are some properties that we can use to simplify some computations.

Exponent Properties

Take $x,y,a$ and $b$ to be real numbers. Then the following properties hold as long as all quantities are defined:

$b^0=1$ where $b\not=0$
$b^x b^y=b^{x+y}$
$\frac{b^x}{b^y}=b^{x-y}$
$\left(ab\right)^x=a^x\cdot b^x$
$\left(\frac{a}{b}\right)^x=\frac{a^x}{b^x}$
$\left(a^x\right)^y=a^{x\cdot y}$
$b^{-x}=\frac{1}{b^x}.$

In the next example, simplify as much as possible using the exponent properties.

Example 2

Assuming that it makes sense, simplify this expression so that the real numbers $x$, $y$, and $z$ are each only written at most one time and the expression is given as a product of monomials:

\[\left(\frac{x^6w^\frac{1}{2}y^{-3}}{y^2x^5}\right)^3.\]

The terms $y^{-3}$ and $y^2$ can be combined like this: \[\frac{y^{-3}}{y^2}=\frac{1}{y^3y^2}=\frac{1}{y^5}.\] Also, the terms $x^{6}$ and $x^5$ can be combined like this: \[\frac{x^6}{x^5}=x^{6-5}=x.\] The entire expression can then be simplified like this: \[\begin{align*} \left(\frac{x^6w^\frac{1}{2}y^{-3}}{y^2x^5}\right)^3&=\left(\frac{xw^\frac{1}{2}}{y^5}\right)^3\\ &=\frac{x^{1\cdot3}w^\frac{3}{2}}{y^{5\cdot 3}}\\ &=\frac{x^3w^\frac{3}{2}}{y^{15}}. \end{align*}\]

The next thing we discuss is the graph of the exponential function. When $b>1$ and $x\geq 0$, the function $\exp_b$ is increasing without bound.

When $b>1$ and $x<0$, then $\exp_b(x)$ is decreasing and approaches zero. So the full graph looks like this:

When $b\in(0,1)$ and $x\geq 0$, $\exp_b(x)$ is decreasing and approaches zero.

When $b\in(0,1)$ and $x< 0$, $\exp_b(x)$ is increasing without bound. So the full graph looks like this.

Regardless of the base, the exponential function has a horizontal asymptote of $y=0.$ However, the behavior depends on the base.

Exponential Function Properties

For $b\in(1,\infty)$:

As $x$ approaches infinity, $b^x$ increases without bound or approaches infinity. Mathematically, we write $x\to \infty$, $b^x\to \infty$.
As $x$ approaches negative infinity, $b^x$ decreases towards zero. Mathematically, we write $x\to -\infty$, $b^x\to 0$.

For $b\in(0,1)$:

As $x$ approaches infinity, $b^x$ decreases towards zero. Mathematically, we write $x\to \infty$, $b^x\to 0$.
As $x$ approaches negative infinity, $b^x$ increases without bound or approaches infinity. Mathematically, we write $x\to -\infty$, $b^x\to \infty$.

Regardless of the base, the exponential function has a horizontal asymptote at $y=0$.

To graph an exponential function, identify its base. That is what determines the shape of its graph.

To graph a function of the form \[g(x)=cb^{ax-h}+k,\] determine the transformations. These kind of functions can be written like this:

\[g=T_k\circ S_c\circ\exp_b\circ S_a\circ T_{-h}.\]

In this case, the transformations that affect the $x$ coordinates of the points are located in the exponent or power of the exponential function. The transformations that affect the $y$ coordinates of the points are located elsewhere.

Practice identifying certain features of an exponential function and graphing it with the following problem.

Example 3

Take $f(x)=5^{x-2}-1.$

Identify the asymptotic behavior of $f$. That is, identify any horizontal or vertical asymptotes.
Determine the domain and range of $f.$
Graph the function $f$ by using transformations on the function $g(x)=5^x.$

The function $f$ can be rewritten in terms of $g$ like this:

\[T_{-1}\circ g\circ T_{-2}.\]

The function $f$ is the exponential function $\exp_5$ that is shifted right 2 and up 1.

The function $\exp_5$ looks like this:

Transform the function so it looks like this:

The asymptotic behavior of $f$ is the horizontal asymptote is $y=-1.$
The domain is $\mathbb{R}$ and the range is $(-1,\infty).$
The graph is given above.

Logarithm Function

The exponential function $\exp_b$, where $b\in(0,1)\cup(1,\infty)$, is an invertible function because for any $y\in(0,\infty)$, the equation \[\exp_b(x)=y\] has a unique solution.

The inverse function has a notation.

Logarithm Function

Take $b\in(0,1)\cup(1,\infty)$. The inverse function $\log_b(x)$ is the logarithm function with base $b$. The relationship is that for any ordered pair of real numbers, $(x,y)$ that satisfy the following exponential equation \[b^{x}=y,\] also satisfy the following logarithm equation \[x=\log_y(x).\]

The domain of the logarithm function is $\mathcal{D}(\log_b)=\mathcal{R}(\exp_b)=(0,\infty).$

The range of the logarithm function is $\mathcal{R}(\log_b)=\mathcal{D}(\exp_b)=(-\infty,\infty).$

The logarithm function has a vertical asymptote at $x=0.$

Another way to see these properties, among others, is to reflect $\exp_b$ along $y=x$:

The graphs of the logarithm look like this:

From here we can see the domain, range, and asymptotic asymptote of the logarithm function.

We will now practice computing the logarithm for certain values. The key is understanding that $\log_b(x)$ will output the power so that $b$ raised to that power equals $x$.

Example 4

Compute the following values:

$\log_2(8)$
$\log_3(27)$
$\log_4(16)$
$\log_\frac{1}{3}(9)$
$\log_\frac{1}{3}\left(\frac{1}{9}\right)$
$\log_2(1)$

$\log_2(8)=\log_2(2^3)=3$
$\log_3(27)=\log_3(3^3)=3$
$\log_4(16)=\log_4(4^2)=2$
$\log_\frac{1}{3}(9)=\log_\frac{1}{3}\left(3^2\right)=\log_\frac{1}{3}\left(\left(\frac{1}{3}\right)^{-2}\right)=-2$
$\log_\frac{1}{3}\left(\frac{1}{9}\right)=\log_\frac{1}{3}\left(\left(\frac{1}{3}\right)^{2}\right)=2$
$\log_2(1)=\log_2(2^0)=0.$

Another way to view the computations we did is by thinking of it as coordinates. For example, $\log_2(8)=3$ means that $(8,3)$ is a point on the logarithm base $2$ function. Reflecting that point over $y=x$ gives the point $(3,8),$ which should be a point on the inverse of the logarithm base $2$ function. It is because this a point on the exponential base $2$ function since $2^{3}=8.$

In viewing it this way, computing something like $\log_3(27)=y$, which can be represented as the point $(27,y)$, can be viewed as finding the right exponent $y$ so that $3^y=27$, which can be represented as the point $(y,27).$

Another way to help with computations is through understanding the special properties of logarithm functions. These properties come from using the properties of the exponential function.

Logarithm Properties

Take $x,y,a$ and $b$ to be real numbers. Then the following properties hold as long as all quantities are defined:

$\log_b(1)=0$ where $b\not=0$
$\log_b(xy)=\log_b(x)+\log_b(y)$
$\log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y)$
$\log_b(b)=1$ where $b\not=0$
$\log_b(a^x)=x\log_b(a)$
$x=\log_b(b^x)$
$x=\exp_b(\log_b(x))$ where $x>0$.

An example of using these properties is by recalculating $\log_4(16):$

\[\begin{align*} \log_4(16)&=\log_4(4^2)\\ &=2\log_4(4)\\ &=2. \end{align*}\]

Practice using the logarithm properties with the next example.

Example 5

Take $x$ and $y$ to be real numbers so that \[\log_3(x)=4\quad\text{and}\quad\log_3(y)=2.\] Compute the following values:

$\log_3(xy)$
$\log_3\left(\frac{x}{y}\right)$
$\log_3\left(x^5\right)$
$\log_3\left(\frac{9y}{x^5}\right)$

Use the logarithm properties for products to get $\log_3(xy)=\log_3(x)+\log_3(y).$ In the problem, it is given that \[\log_3(x)=4\quad\text{and}\quad\log_3(y)=2\] so we get that \[\begin{align*} \log_3(xy)&=\log_3(x)+\log_3(y)\\ &=4+2\\ &=6. \end{align*}\]
Use the logarithm properties for quotients to get $\log_3\left(\frac{x}{y}\right)=\log_3(x)-\log_3(y).$ In the problem, it is given that \[\log_3(x)=4\quad\text{and}\quad\log_3(y)=2\] so we get that \[\begin{align*} \log_3(xy)&=\log_3(x)-\log_3(y)\\ &=4-2\\ &=4. \end{align*}\]
Use the logarithm properties for exponents to get $\log_3\left(x^5\right)=5\log_3(x).$ In the problem, it is given that \[\log_3(x)=4\] so we get that \[\begin{align*} \log_3\left(x^5\right)&=5\log_3(x)\\ &=5\cdot 4\\ &=20. \end{align*}\]
Use the logarithm properties for quotients to get $\log_3\left(\frac{9y}{x^5}\right)=\log_3(9y)-\log_3(x^5).$ Use the logarithm properties for products and exponents to get that \[\log_3(9y)=\log_3(9)+\log_3(y)\quad\text{and}\quad\log_3(x^5)=5\log_3(x).\] Use what was given in the problem to obtain that \[\begin{align*} \log_3\left(\frac{9y}{x^5}\right)&=\log_3(9y)-\log_3(x^5)\\ &=\log_3(9)&&+&&&\log_3(y)&&&&-&&&&&5\log_3(x)\\ &=2&&+&&&2&&&&-&&&&&5\cdot 4\\ &=-16. \end{align*}\]

To graph a logarithm function, identify its base. Just like with the exponential function, its base determines the shape of its graph.

To graph a function of the form \[g(x)=c\log_b(ax-h)+k,\] determine the transformations. These kind of functions can be written like this:

\[g=T_k\circ S_c\circ\log_b\circ S_a\circ T_{-h}.\]

In this case, the transformations that affect the $x$ coordinates of the points are located inside of the logarithm function. The transformations that affect the $y$ coordinates of the points are located elsewhere.

Practice identifying certain features of a logarithm function and graphing it with the following problem.

Example 6

Take $f(x)=3\log_\frac{1}{2}(-x+1).$

Identify the asymptotic behavior of $f$. That is, identify any horizontal or vertical asymptotes.
Determine the domain and range of $f.$
Graph the function $f$ by using transformations on the function $g(x)=\log_\frac{1}{2}(x).$

The function $f$ can be rewritten in terms of $g$ like this:

\[S_{3}\circ g\circ S_{-1}\circ T_{1}.\]

The function $f$ is the logarithm function $\log_\frac{1}{2}$ that is shifted left 1, scaled in the $x$ coordinate by $-1$ (or reflected along $y$-axis), and then scaled in the $y$ coordinate by $3$.

The function $\log_\frac{1}{2}$ looks like this:

Transform the function so it looks like this:

The asymptotic behavior of $f$ is the vertical asymptote is $x=1.$
The domain is $(-\infty,1)$ and the range is $\mathbb{R}.$
The graph is given above.

Exponential Base Euler’s Number Function and Natural Logarithm Function

Here we make a quick, but important, note on two important things. One is an important constant in mathematics. The other note is an exponential and logarithm function with this constant as its base. All properties we discussed before still hold, but the notation used to describe these two functions are slightly different.

Euler’s Number, Exponential and Natural Logarithm Function

The constant $e$ is known as Euler’s number. It is this irrational number: \[e=2.7182818\dots\] There are two functions defined with this number as its base.

The exponential function is used to refer to the exponential function with base $e$. It is written as $\exp(x)=\exp_e(x).$
The natural logarithm function is used to refer to the logarithm function with base $e$. It is written as $\ln(x)=\log_e(x).$

Because the base $e$ is in the interval $(1,\infty)$, $\exp$ and $\ln$ both behave like $\exp_b$ and $\log_b$ where $b$ is in the interval $(1,\infty).$ As a result, graphing these functions can be done just like we did in the previous problems.

Practice some computations with this problem. The important thing to remember is that $\ln(x)$ is $\log_e(x)$, so try to rewrite the input as a power of $e$, whenever possible.

Example 7

Compute the following values:

$\ln(1)$
$\ln(e)$
$\ln(e^4)$
$\ln\left(\frac{1}{e}\right)$
$\ln(4)$

$\ln(1)=\ln(e^0)=0$
$\ln(e)=\ln(e^1)=1$
$\ln(e^4)=4$
$\ln\left(\frac{1}{e}\right)=\ln\left(e^{-1}\right)=-1$
There is no “nice” number (rational number), $x$, so that $4=e^x$. And so $\ln(4)$ must be left as is. It is a real number.

Exponential and Logarithm Equations

Solving equations involving logarithms and exponential terms can be solved if they can be rewritten as

\[\exp_b(X)=\exp_b(Y)\quad\text{or}\quad\log_b(X)=\log_b(Y)\]

where $X$ and $Y$ are algebraic expressions. This is because of a special property these two functions possess, which is that the equations can only hold if $X=Y$. Of course, there is only a solution as long as all expressions make sense, so it is important verify there are no issues with domain.

Practice solving the following equations. It will be helpful to remember the other properties an exponential function and logarithm function possess to help with rewriting them into one of the two forms listed above.

Example 8

Solve the following equations:

$5^{x-1}-25^{x}=0$
$e^{x}-4=0$
$\log_3(x)+\log_3(x+4)=1.$

The equation can be rewritten like this: $5^{x-1}=25^{x}.$ The number $25$ can be rewritten like this: \[25=5^2.\] And so \[\begin{align*}5^{x-1}&=25^x\\5^{x-1}&=\left(5^{2}\right)^x\\5^{x-1}&=5^{2x}\\x-1&=2x, \end{align*}\]where the property $5^{X}=5^{Y}$ implies that $X=Y$ is used. Solve the equation $x-1=2x$ to get that $x=-1$ is the solution.
The equation can be rewritten like this: $e^{x}=4.$ The number $4$ can be rewritten like this: \[4=e^{\ln(4)}.\] And so \[\begin{align*}e^{x}&=4\\e^{x}&=e^{\ln(4)}\\x&=\ln(4), \end{align*}\] where the property $e^{X}=e^{Y}$ implies that $X=Y$ is used. The solution is thus $x=\ln(4).$
Notice that we have logarithm terms, so the solution must be a number in the domain of both $\log_3(x)$ and $\log_3(x+4)$. For the first logarithm function, its domain is $(0,\infty)$. For the second logarithm function, its domain is $(-4,\infty)$. The intersection of these two sets is $(0,\infty),$ so we are looking on for solutions in this interval. Use the logarithm properties to rewrite the left term like this: \[\log_3(x)+\log_3(x+4)=\log_3(x(x+4)).\] The right term can be rewritten like this: \[1=\log_3(3^1).\] And so \[\begin{align*}\log_3(x)+\log_3(x+4)&=1\\ \log_3(x(x+4))&=\log_3(3^1)\\x(x+4)&=3, \end{align*}\] where the property $\log_3(X)=\log_3(Y)$ implies that $X=Y$ is used. Solve the equation $x(x+4)=3$, which is equivalent $x^2+4x-3=0.$ The solution to the quadrant equation is $x=\sqrt{7}-2$ and $x=-\sqrt{7}-2.$ However, the solution $x=-\sqrt{7}-2\approx -4.64$ is not in the intersection of the domain of $(0,\infty).$ So the only solution is therefore, $x=\sqrt{7}-2\approx 0.64.$

Modeling Growth

The final thing we study is models of change. We focus on models of change that have a great deal of symmetry.

To study change, we focus on change in the following way: rather than focus on inputs a distance of $1$ away, we focus on intervals of the same length.

Specifically, on any two intervals of the same length, the models we study change by either

the same amount,
by the same factor.

In the case of where the change is by the same amount, the change will be linearly. Specifically, if the amount of the change is a real number $M$, then for any real number $t$, the function $f$ looks like this: \[f(t)=f(0)+Mt.\]

In the following example, we create a formula for a model that observes linear change given a data set.

Example 9

A quantity $A$ experiences linear change and has the following data set: \[\begin{cases} A(0)&=15\\ A(1)&=29. \end{cases}\] Write a model for $A$.

The model looks like this for some number $M$: \[\begin{align*} A(t)&=A(0)+Mt\\ &=15+Mt. \end{align*}\] Use the information that $A(1)=29$ and the above model to obtain that: \[\begin{align*} A(t)&=15+Mt\\ A(1)&=15+M\cdot 1\\ 29&=15+M\\ 14&=M. \end{align*}\]

The model is thus

\[A(t)=15+14t.\]

In the next example, we will work it by using rigidity. Knowing that the form of the linear model will have to be of the form, for some constants $C,D,a$ and $b$,

\[A(t)=C+D\frac{t-a}{b}.\]

Example 10

A quantity $A$ experiences linear change and has the following data set: \[\begin{cases} A(2)&=35\\ A(7)&=65. \end{cases}\] Write a model for $A$.

Because the model experiences linear change, the linear change $M$ will be \[M=65-35=30.\] Since the model requires that $A(2)=35$ it must look something like this: \[\begin{align*} A(t)&=35+M\frac{t-2}{b}\\ &=35+30\frac{t-2}{b}. \end{align*}\] Use the information that $A(7)=65$ and the above model to obtain that: \[\begin{align*} A(t)&=35+30\frac{t-2}{7-2}\\ &=35+30\frac{t-2}{5}\\ &=35+6(t-2). \end{align*}\]

The model is thus

\[A(t)=15+14t.\]

In the case of where the change is by the same factor, the change will be exponential. Specifically, if the factor is a real number $b$, then for any real number $t$, the function $f$ looks like this: \[f(t)=f(0)b^t.\]

In the following example, we create a formula for a model that observes exponential change given a data set.

Example 11

A quantity $A$ experiences exponential change and has the following data set: \[\begin{cases} A(0)&=4\\ A(1)&=9. \end{cases}\] Write a model for $A$.

The model looks like this for some number $b$: \[\begin{align*} A(t)&=A(0)b^t\\ &=4\cdot b^t. \end{align*}\] Use the information that $A(1)=9$ and the above model to obtain that: \[\begin{align*} A(t)&=4\cdot b^t\\ A(1)&=4\cdot b^{1}\\ 9&=4\cdot b\\ \frac{9}{4}&=b. \end{align*}\]

The model is thus

\[A(t)=4\cdot \left(\frac{9}{4}\right)^t.\]

In the next example, we will work it by using rigidity. Knowing that the form of the linear model will have to be of the form, for some constants $C,D,a$ and $b$,

\[A(t)=C\cdot b^\frac{t-a}{D}.\]

Example 12

A quantity $A$ experiences exponential change and has the following data set: \[\begin{cases} A(8)&=6\\ A(12)&=15. \end{cases}\] Write a model for $A$.

Because the model experiences exponential change, and we need $A(8)=6$, it will look something like this \[\begin{align*} A(t)&=6\cdot b^\frac{t-8}{?}.\\ \end{align*}\]

It needs to be the case that at $t=12$, $A(12)=15$ so $b=\frac{15}{6}$ to get the right number. Scale the $t$ as well to get

\[\begin{align*} A(t)&=6\cdot b^\frac{t-8}{?}.\\ &=6\cdot\left(\frac{15}{6}\right)^\frac{t-8}{12-8}\\ &=6\cdot\left(\frac{15}{6}\right)^\frac{t-8}{4}. \end{align*}\]

The model is thus

\[A(t)=8\cdot\left(\frac{15}{6}\right)^\frac{t-8}{4}.\]

When a quantity experiences unrestricted growth, it is an example of exponential growth. And so, it is modeled using the exponential function. The model looks like this:

\[A(t)=A(0)e^{kt}\] where $k$ is a positive constant known as the growth factor.

Getting the growth factor of an exponential growth model involves rewriting the model so that it looks like the above form. Practice that with the next example.

Example 13

A mass of bacteria experiences exponential growth. At time $5$, there is $15$ grams of the bacteria. At time $9$, the mass has grown to $28$.

Determine the growth rate of the substance.
Determine the doubling time of the bacteria.
Determine the time that it takes for amount of the bacteria to increase by a factor of $3$

Because the model experiences exponential change, and we need $A(5)=15$, it will look something like this \[\begin{align*} A(t)&=15\cdot b^\frac{t-5}{?}.\\ \end{align*}\]

It needs to be the case that at $t=9$, $A(9)=28$ so $b=\frac{28}{15}$ to get the right number. Scale the $t$ as well to get

\[\begin{align*} A(t)&=15\cdot b^\frac{t-5}{?}.\\ &=15\cdot \left(\frac{28}{15}\right)^\frac{t-5}{9-5}\\ &=15\cdot\left(\frac{28}{15}\right)^\frac{t-5}{4}. \end{align*}\]

The model is thus

\[A(t)=15\cdot\left(\frac{28}{15}\right)^\frac{t-5}{4}.\]

However, to determine the growth factor, we need to rewrite it so the model looks like this

\[A(t)=A(0)e^{kt}.\]

First split up the exponential term using exponent properties:

\[\left(\frac{28}{15}\right)^\frac{t-5}{4}=\left(\frac{28}{15}\right)^{\frac{1}{4}t}\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}.\]

Now rewrite the $\frac{28}{15}$ term like this: \[\frac{28}{15}=e^{\ln(\frac{28}{15})}.\]

Combine everything to get that the model looks like this: \[\begin{align*} A(t)&=15\cdot\left(\frac{28}{15}\right)^\frac{t-5}{4}\\ &=15\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}\cdot \left(\frac{28}{15}\right)^{\frac{1}{4}t}\\ &=15\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}\cdot e^{\ln(\frac{28}{15})\cdot\frac{1}{4}t}\\ &=15\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}\cdot e^{\frac{\ln\left(\frac{28}{15}\right)}{4}t}\\ \end{align*}\]

The growth rate is $k=\frac{\ln\left(\frac{28}{15}\right)}{4}.$
To find the doubling time, we need to find a time $T$ so that the model looks like this: \[A(t)=A(0)2^\frac{t}{T}.\] We already have \[A(t)=\left(15\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}\right)\cdot e^{\frac{\ln\left(\frac{28}{15}\right)}{4}t}.\] The equality of the two models only holds if \[e^{\frac{\ln\left(\frac{28}{15}\right)}{4}}=2^\frac{1}{T}.\] Solve for $T$ to get \[\begin{align*} e^{\frac{\ln\left(\frac{28}{15}\right)}{4}}&=\left(2\right)^\frac{1}{T}\\ \frac{\ln\left(\frac{28}{15}\right)}{4}&=\ln\left(2\right)^\frac{1}{T}\\ \frac{\ln\left(\frac{28}{15}\right)}{4}&=\frac{1}{T}\ln\left(2\right)\\ T&=\frac{4\ln\left(2\right)}{\ln\left(\frac{28}{15}\right)}. \end{align*}\] The doubling time is thus $T=\frac{4\ln\left(2\right)}{\ln\left(\frac{28}{15}\right)}.$
To find the time, we need to find a time $T$ so that the model looks like this: \[A(t)=A(0)\left(3\right)^\frac{t}{T}.\] We already have \[A(t)=\left(15\cdot \left(\frac{28}{15}\right)^\frac{-5}{4}\right)\cdot e^{\frac{\ln\left(\frac{28}{15}\right)}{4}t}.\] The equality of the two models only holds if \[e^{\frac{\ln\left(\frac{28}{15}\right)}{4}}=3^\frac{1}{T}.\] Solve for $T$ to get $T=\frac{4\ln\left(3\right)}{\ln\left(\frac{28}{15}\right)}.$

When a quantity experiences radioactive decay, it is an example of exponential decay. And so, it is modeled using the exponential function. The model looks like this:

\[A(t)=A(0)e^{-kt}\] where $k$ is a positive constant known as the decay factor.

Getting the decay factor of an exponential decay model involves rewriting the model so that it looks like the above form. Practice that with the next example.

Example 14

There is a radioactive substance such that at time $2$, there is $24$ grams of it. At time $13$, there remains $7$ grams.

Determine the decay rate of the substance.
Determine the half-life of the substance.
Determine the time that it takes for only a fraction of $\frac{1}{3}$ of the original remains.

Because the model experiences exponential change, and we need $A(2)=24$, it will look something like this \[\begin{align*} A(t)&=24\cdot b^\frac{t-2}{?}.\\ \end{align*}\]

It needs to be the case that at $t=13$, $A(13)=7$ so $b=\frac{7}{24}$ to get the right number. Scale the $t$ as well to get

\[\begin{align*} A(t)&=24\cdot b^\frac{t-2}{?}.\\ &=24\cdot \left(\frac{7}{24}\right)^\frac{t-2}{13-2}\\ &=24\cdot\left(\frac{7}{24}\right)^\frac{t-2}{11}. \end{align*}\]

The model is thus

\[A(t)=24\cdot\left(\frac{7}{24}\right)^\frac{t-2}{11}.\]

However, to determine the growth factor, we need to rewrite it so the model looks like this

\[A(t)=A(0)e^{-kt}.\]

First split up the exponential term using exponent properties:

\[\left(\frac{7}{24}\right)^\frac{t-2}{11}=\left(\frac{7}{24}\right)^{\frac{1}{11}t}\cdot \left(\frac{7}{24}\right)^\frac{-2}{11}.\]

Now rewrite the $\frac{7}{24}$ term like this: \[\frac{7}{24}=e^{\ln(\frac{7}{24})}.\]

Combine everything to get that the model looks like this: \[\begin{align*} A(t)&=24\cdot\left(\frac{7}{24}\right)^\frac{t-2}{11}\\ &=24\cdot \left(\frac{7}{24}\right)^\frac{-2}{11}\cdot e^{\frac{\ln\left(\frac{7}{24}\right)}{11}t}\\ \end{align*}\]

The decay rate is $k=-\frac{\ln\left(\frac{7}{24}\right)}{11}.$
To find the half-life, we need to find a time $T$ so that the model looks like this: \[A(t)=A(0)\left(\frac{1}{2}\right)^\frac{t}{T}.\] We already have \[A(t)=\left(24\cdot \left(\frac{7}{24}\right)^\frac{-2}{11}\right)\cdot e^{\frac{\ln\left(\frac{7}{24}\right)}{11}t}.\] The equality of the two models only holds if \[e^{\frac{\ln\left(\frac{7}{24}\right)}{11}}=\left(\frac{1}{2}\right)^\frac{1}{T}.\] Solve for $T$ to get that the half-life is $T=\frac{11\ln\left(\frac{1}{2}\right)}{\ln\left(\frac{7}{24}\right)}.$
To find the time, we need to find a time $T$ so that the model looks like this: \[A(t)=A(0)\left(\frac{1}{3}\right)^\frac{t}{T}.\] We already have \[A(t)=\left(24\cdot \left(\frac{7}{24}\right)^\frac{-2}{11}\right)\cdot e^{\frac{\ln\left(\frac{7}{24}\right)}{11}t}.\] The equality of the two models only holds if \[e^{\frac{\ln\left(\frac{7}{24}\right)}{11}}=\left(\frac{1}{3}\right)^\frac{1}{T}.\] Solve for $T$ to get $T=\frac{11\ln\left(\frac{1}{3}\right)}{\ln\left(\frac{7}{24}\right)}.$

Return

Return

\(x\)	\(g(x)\)
\(0\)	\(1\)
\(1\)	\(2\)
\(2\)	\(4\)
\(3\)	\(8\)
\(4\)	\(16\)