]> Chapter 04 Notes (Sullivan & Sullivan 3e)

Chapter 04: Exponential and Logarithmic Functions

§4.1: One-to-One Functions; Inverse Functions

One-to-One Functions

In order for a relation to be a function, it has to pass the Vertical Line Test—in other words, for each value of x there is exactly one value of y. There is a symmetric analogue to this…when every y value has exactly one value of x (in addition to every x having exactly one y value), then we say that the function is one-to-one. If you draw one of those mapping diagrams, there will be a bunch of segments that connect points in the domain to points in the range, so that every point is on exactly one segment.

Visually, a one-to-one function will pass both the Vertical Line Test and the Horizontal Line Test—every horizontal line will intersect the graph at no more than one point.

Inverse Functions

One result of this one-to-one relationship between the domain and range is that it is easy to use a function both forwards and backwards. Think of this: for the function $f (x) = x^{2}$ , it is easy to plug in a value of x and get the value of y ( $f (2) = 4$ ), but it isn’t as easy to start with the y value and work back to the x (if $f (x) = 16$ , then what’s x?).

One-to-one functions will have Inverse Functions—functions that work backwards. For example, the function $g (x) = 2 x + 1$ has the inverse function $g^{- 1} (x) = \frac{x - 1}{2}$ . Notice the notation: the exponent on the name of the function does not mean reciprocal! In other words, $g^{- 1} (x) \neq \frac{1}{g (x)}$ . An inverse function will “undo” whatever the original function did to an input…with the most recent example, $g (4) = 9$ , so $g^{- 1} (9) = 4$ . The domain of the function will be the range of the inverse function, and the range of the original function will be the domain of the inverse function. This also means that the graph of $y = f^{- 1} (x)$ will be the reflection of $y = f (x)$ across the line $y = x$ .

Finding the rule for an inverse function is easy: swap the symbols x and y, and then solve for y. Note that you may occasionally have to specify a domain in order to really have a true inverse function.

Examples

[1.] Determine if $h (x) = x^{4} + 2 x^{3} - 3 x^{2} - 4 x + 4$ is one-to-one.

Let’s look at a graph:

An Image

Since this does not pass the Horizontal Line Test, it is not one-to-one.

[2.] Find the inverse function for $f (x) = \frac{2 x + 2}{3 x - 1}$ .

First of all, does this even have one?

An Image

Looks like it does.

Start by swapping x and y: $x = \frac{2 y + 2}{3 y - 1}$ . Now, solve for y: $3 x y - x = 2 y + 2 \Rightarrow 3 x y - 2 y = x + 2 \Rightarrow y (3 x - 2) = x + 2 \Rightarrow y = \frac{x + 2}{3 x - 2}$ . Thus, $f^{- 1} (x) = \frac{x + 2}{3 x - 2}$ . If you can solve for y without a plus or minus symbol (or an absolute value), then it’s a good bet that you’ve found an inverse function.

[3.] Find the inverse function of $q (x) = \sqrt{x}$ .

I know what that looks like, so I know it has an inverse function. $x = \sqrt{y} \Rightarrow x^{2} = y$ . However, since the domain and range of $q (x)$ are $[0, \infty)$ , I have to specify that $q^{- 1} (x) = x^{2}$ only for $x \in [0, \infty)$ .

§4.2: Exponential Functions

Reminder: Rules of Exponents

You should know these!

$a^{m} \cdot a^{n} = a^{m + n}$

$\frac{a^{m}}{a^{n}} = a^{m - n}$

${(a^{m})}^{n} = a^{m n}$

$a^{0} = 1$

$a^{- 1} = \frac{1}{a}$

Exponential Functions

An exponential function is a function of the form $f (x) = n^{x}$ where $n \in ℝ^{+}$ (we can’t allow non-positive bases…for reasons related to the next parenthetical statement). The domain of this parent function is all real numbers (though you don’t yet truly understand what something like $2^{3.1}$ really means…) and the range is $y \in (0, \infty)$ .

Graphs of Exponential Functions

It should not be hard to see that an exponential function must pass through the points $(0, 1)$ and $(1, n)$ . Depending on the value of n, there will be a horizontal asymptote on one side of the function, and the other side will run up towards infinity. The fact that there is an asymptote on only one side should not bother you…we’re already said that horizontal asymptotes only matter at the ends; it’s not a big leap to see that it might only matter on one end.

Note that if $n > 1$ , the function will always be increasing, and if $n \in (0, 1)$ that the function will always be decreasing.

Once you’ve mastered the graph of the parent function, then you can tackle any transformation: $y = a \cdot n^{b (x - c)} + d$ .

Solving Exponential Equations

For now, we’ll solve these graphically.

Examples

[4.] Sketch $y = 1 - 2 \cdot 3^{x + 1}$ .

Take the parent graph of $y = 3^{x}$ , flip it upside down and multiply the y-values by 3, shift it one left and up one.

An Image

[5.] Solve $2^{3 x - 5} = 64$ .

I don’t need a graph for this! I’ll make 64 a base of 2 and go from there. $2^{3 x - 5} = 2^{6} \Rightarrow 3 x - 5 = 6 \Rightarrow 3 x = 11 \Rightarrow x = \frac{11}{3}$ .

[6.] Solve $e^{x - 3} = x^{2}$ .

This time I’ll solve graphically. I’ll actually graph $y = e^{x - 3} - x^{2}$ and look for x-intercepts.

An Image

The solutions are -0.201, 0.253 and 6.848.

§4.3: Logarithmic Functions

The Definition of a Logarithm

Note that exponential functions are one-to-one; thus, they must have inverse functions.

For the exponential function $y = n^{x}$ , the inverse function is $y = \log_{n} (x)$ (that’s read “the log of x, base n). The symbol log is a function which must act on some input; writing it by itself is meaningless. Writing it without a base is ambiguous; in high school, writing it without a base means base 10 (Common Logarithm). Writing log without a base outside of high school means base e (Natural Logarithm). I don’t personally like to do unnatural things, so when I say “log” I mean “natural log.” I’ll write ln for natural log, though.

What’s that? You don’t know Euler’s Number? Euler’s number e is the horizontal asymptote of the function $y = {(1 + \frac{1}{x})}^{x}$ .

Note that if $q = n^{t}$ , then $\log_{n} (q) = t$ . You should be able to use this to convert expressions between exponential and logarithmic forms.

Logarithmic Functions

Because exponentials and logarithms are inverse functions, their domains and ranges are reversed—thus, the domain of a logarithmic function is $x \in (0, \infty)$ and the range is $y \in ℝ$ . The restriction on the value of n is the same: $n \in ℝ^{+}$ .

Graphs of Logarithmic Functions

The exponential function $y = n^{x}$ passes through the points $(0, 1)$ and $(1, n)$ ; thus, the logarithmic function $y = \log_{n} (x)$ passes through $(1, 0)$ and $(n, 1)$ . Where the exponential function has a horizontal asymptote at $y = 0$ , the logarithmic function will have a vertical asymptote at $x = 0$ . The actual graph of $y = \log_{n} (x)$ can be found by flipping the graph of $y = n^{x}$ across the line $y = x$ .

Once you know what the parent graph looks like, then you should be able to graph any transformation of it: $y = a \cdot \log_{n} (b (x - c)) + d$ .

Solving Logarithmic Equations

For now, we’ll solve logarithmic equations by either graphing, or by switching back to exponential form.

Examples

[7.] Convert to logarithmic form: $2^{x + 1} = 8$ .

$\log_{2} (8) = x + 1$

[8.] Evaluate: $\log_{5} (125)$ .

Since $5^{3} = 125$ , $\log_{5} (125) = 3$ .

[9.] Sketch the graph of $y = \log_{2} (x - 4)$ .

Take the parent graph of $y = \log_{2} (x)$ —which passes through $(1, 0)$ and $(1, 2)$ —and shift it right 4 units.

An Image

[10.] Solve $\log_{x} (15) = 2$ .

I’ll convert back to exponential form and solve: $x^{2} = 15 \Rightarrow x = \pm \sqrt{15}$ . However, since x was in the base of the logarithm, I can’t allow a negative answer; thus, the solution is $x = \sqrt{15}$ .

§4.4: Properties of Logarithms

The Properties

These properties can be derived from the rules of exponents…

$\log_{n} (a \cdot b) = \log_{n} (a) + \log_{n} (b)$

$\log_{n} (\frac{a}{b}) = \log_{n} (a) - \log_{n} (b)$

$\log_{n} (a^{b}) = b \cdot \log_{n} (a)$

The book mentions two more that I have previously mentioned as known points on the graph…

The Change of Base Formula

The only two bases on your calculator are base 10 and base e…in many computer programs, the only base is e! Thus, you may need to employ the change of base formula in order to evaluate certain logarithms.

For any valid base n, $\log_{b} (a) = \frac{\log_{n} (a)}{\log_{n} (b)}$ .

Examples

[11.] Expand the expression to a sum or difference of logarithms: $\ln (3 {(x - 2)}^{4})$ .

$\ln (3 {(x - 2)}^{4}) = \ln (3) + \ln {(x - 2)}^{4} = \ln (3) + 4 \ln (x - 2)$

[12.] Collapse the expression to a single logarithm: $2 \log (x) - 3 \log (x - 2)$ .

$2 \log (x) - 3 \log (x - 2) = \log (x^{2}) - \log ({(x - 2)}^{3}) = \log (\frac{x^{2}}{{(x - 2)}^{3}})$

[13.] Evaluate $\log_{3} (21)$ .

I’ll enter $\frac{\ln (21)}{\ln (3)}$ in my calculator to get 2.771.

§4.5: Logarithmic and Exponential Equations

If needed, you could take the logarithm of each side of the equation, or you could raise each side as a power of some base…maybe you might even need to use the properties of exponents or logarithms…

Examples

[14.] Solve $e^{2 x - 1} = 7$ .

I’ll take the natural log of both sides: $e^{2 x - 1} = 7 \Rightarrow 2 x - 1 = \ln (7) \Rightarrow 2 x = \ln (7) + 1 \Rightarrow x = \frac{\ln (7) + 1}{2}$ . If you had a calculator you could get an approximate value for that expression…but then you would have solved the whole thing graphically, right?

[15.] Solve $\log (x - 8) + \log (x + 8) = 2$ .

$\log (x - 8) + \log (x + 8) = 2 \Rightarrow \log (x^{2} - 64) = 2 \Rightarrow x^{2} - 64 = 100 \Rightarrow x^{2} = 164 \Rightarrow x = \pm \sqrt{164}$ . Note that this is very close to $\pm 13$ ; thus, the $x - 8$ part would be negative, and you can’t take the log of a negative number…thus, the answer is $x = \sqrt{164}$ .

§4.6: Compound Interest

One application of exponentials is interest…where you either receive (or pay) extra for the use of money (by either giving it to the bank, or borrowing it from someone).

Simple Interest

Simple Interest means that interest is only calculated on the Principal (the initial amount of money). In this case, $V (t) = P + P \cdot r \cdot t = P (1 + r \cdot t)$ , where P is the principal, r is the rate of interest and t is the amount of time elapsed. It is usually the case that r is given as an Annual Percentage Rate (also know as Per Annum) and that t is given in years…however, this formula works as long as the time period for r is the same as the time period for t.

Compound Interest

Compound Interest means that interest is calculated on both the principal and the interest earned. In this case, $V (t) = P {(1 + \frac{r}{k})}^{k t}$ . Once again, P is the principal…but for this formula, r is the annual rate of interest. k is the number of interest periods per year, and t is the number of years.

As k approaches infinity, this formula morphs into the Continuous Interest Formula: $V (t) = P e^{r t}$ .

The Effective Rate of Interest is rate needed to make the value after compound interest equal to the value after simple interest (after one year). It is not hard to set up an equation to solve for this rate…

Examples

[16.] Find the value of $2000 invested in an account that earns 3% APR (compounded monthly) for five years.

$V (5) = 2000 {(1 + \frac{0.03}{12})}^{12 \cdot 5} \approx 2323.23356$ . The investment will be worth approximately $2323.23.

[17.] How much money must be invested at 2% continuous interest in order to end up with $10,000 after ten years?

Solve: $10000 = P e^{0.02 \cdot 10} \Rightarrow P = \frac{10000}{e^{0.2}} \approx 8187.3075$ . The principal would have been about $8187.31.

[18.] What is the effective rate of an account that compounds a 1% APR monthly?

Solve: $P {(1 + \frac{0.01}{12})}^{12} = P (1 + r) \Rightarrow r = {(1 + \frac{0.01}{12})}^{12} - 1 \approx 0.01004596$ . The effective rate is 1.005%.

§4.7: Growth and Decay

Another application of exponentials is growth and decay of certain populations…

Exponential Growth and Decay

If a population can grow without bound (unlikely!) or decay to zero (more likely), then the population at any time can be modeled by $P (t) = P_{0} {(1 + r)}^{t}$ . In this equation, $P_{0}$ is the initial population, r is the rate of growth or decay, and t is the time (using the same time units as r).

One common version of this formula is used to model the decay of radioactive substances, where we often refer to the half-life of a radioactive substance (the time required for half of the mass to decay and radiate as energy)…in this application, the formula is often written $P (t) = P_{0} {(\frac{1}{2})}^{\frac{t}{k}}$ , where k is the half-life of the substance and t is time (in units that match k).

Another version of this formula is $P (t) = P_{0} e^{k t}$ , where k is a constant related to the rate of growth or decay…the choice of e as the base is related to Calculus (it’s the only base that we really understand!).

Logistic Growth

When it comes to growth, it’s really unusual for a population to grow without bound—there is almost always some natural upper limit to that growth. A way to model this is with a Logistic Model: $P (t) = \frac{P_{\infty}}{1 + k e^{- r t}}$ , where $P_{\infty}$ is the upper limit on the growth, r is the rate of growth, and k is a constant related to the growth. This function has two horizontal asymptotes—one at zero and one at $P_{\infty}$ .

Examples

[19.] The number of bacteria in a culture triple in number every ten hours. If there are 900 bacteria initially, then how long will it take for there to be 5000 bacteria?

Note that tripling is a 200% increase. The bacteria population model is $P (t) = 900 {(3)}^{\frac{t}{10}}$ . We need to solve $5000 = 900 {(3)}^{\frac{t}{10}}$ . Easy enough! $\frac{50}{9} = 3^{\frac{t}{10}} \Rightarrow \ln (\frac{50}{9}) = (\frac{t}{10}) \ln (3) \Rightarrow t = \frac{10 \ln (\frac{50}{9})}{\ln (3)} \approx 15.609$ . It should take about 15.609 hours for the bacteria to reach a population size of 5000.

[20.] How long will it take for a 10g sample of fluorine-18 to reduce to less than 1g? The half life of fluorine-18 is 1.82 hours.

The population model is $A (t) = 5 {(\frac{1}{2})}^{\frac{t}{1.82}}$ and we need to solve $1 = 5 {(\frac{1}{2})}^{\frac{t}{1.82}}$ …I’ll do this one graphically by graphing $y = 5 {(0.5)}^{\frac{t}{1.82}} - 1$ and looking for x-intercepts.

An Image

It takes 4.226 hours for 5g to decay to 1g.

[21.] The infection rate (number of new cases) for a particular disease is given by the equation $I (t) = \frac{50000}{1 + e^{5 - 0.8 t}}$ (where t is the number of years after the first recorded cases). How much time did it take for the infection rate to double from 20,000 per year to 40,000 per year?

Let’s graph this.

An Image

The rate is at 20,000 at time 5.743 years, and the rate hit 40,000 at time 7.983 years, so it took 2.24 years for the rate to double.

§4.8: Models

We’re not going to do this…forget you saw it.

Page last validated 2010-08-15