**Here's a riddle**: I'm offering you a job. It's a nasty one, like mucking out chicken coops (big ones, hundreds of chickens – and "mucking out" means shoveling ammonia-smelling chicken poop) 12 hours a day, for 30 days — no lunch breaks. At the end of the first day, I'll pay you one cent. **One cent**. At the end of the second, I'll double that and give you two cents, and so on for 30 days.

Think it through. On the seventh day, I'll pay you 64 cents, and you will have done 84 total hours of wretched labor for a grand total of \$1.27. Congratulations on your great decision and awesome survival instinct. Ready to quit yet?

Skip ahead to day 14. At the end of the day, I'll pay you \$81.92. Now that's more like it. And your cumulative total is now up to \$163.83 ... better, and remember that tomorrow your wage will double to \$163.84 for one day's work. Would you take the job?

In the table are the daily payments and cumulative earnings for just the** last 10 days** of your employment in the coops.

On the last dayalone, I'd have to pay you over 5 million dollars! All-told, you'd earn more than \$10.7 million! Take the job!

That's **exponential growth**, and you'll see it in many different phenomena, like financial growth (**compound interest**) and **population growth**. Its cousin, **exponential decay** is used to model things like radioactive decay, the decrease in drug concentration in the bloodstream over time and the depreciation (loss of value) of property over time. And there are many other examples.

Knowing how to manipulate and solve exponential functions and equations is a very important part of your mathematics toolkit.

The most important feature of an exponential function is that **the independent variable is in the exponent** of some base, usually an agreed-upon (but constant) base, like 2, 10 or the special number "*e*".

The graphs of exponential functions are characterized by a period of relatively slow growth followed by much more rapid growth, overtaking polynomial functions of even the highest degree.

where **b** is the base and the independent variable (**x**) is *in the exponent*.

Generally, it doesn't make any sense to have a negative base in an exponential function (b < 0). That's because, for integer exponents, the sign of the function will alternate. It's even worse for rational exponents, which represent roots which may or may not exist. If we want a negative-valued exponential function, we can always write something like

$$f(x) = -b^x,$$

which is *not* the same as $(-b)^x.$

One notable exception is that in some sums, we use the term $(-1)^n,$ where n is an integer, to purposefully cause alternate values of a sum to alternate in sign. You can read more about that in the section on alternating series.

In an **exponential function**, the variable is in the exponent. The base is constant.

In this illustration, the functions $f(x) = x^2$ (

Notice the difference in the two functions when $x \lt 0.$ Recall from the laws of exponents that $a^{-1}$ means to take the reciprocal of **a**. So as **x** grows in the **negative** direction, the function asymptotically approaches zero: **½, ⅓, ¼**, ...

The value of an exponential functions is never less than zero unless the function is specifically multiplied by a negative number - a vertical scaling parameter. Remember that a negative exponent does not make the *number* negative: $a^{-1} = 1/a.$

Exponential graphs all have a common shape, and sketching them is pretty easy. Check out the graph of $f(x) = 2^x$ on the right.

Without any transformations (covered below), every exponential function $f(x) = b^x,$ where **b** is a constant base (2 in this case), passes through **y = 1**, because anything raised to the zero power is 1: **b ^{0} = 1**

On the right we have the rapid exponential growth. On the left, as the exponent approaches -∞, the function approaches zero. Remember that a negative exponent like b^{-1} doesn't mean a negative number; it means 1/b, so as x → ∞, the value of the function approaches zero, thus we have asymptotic behavior on the left.

$f(x) = b^{-x}$ does not mean that f(x) is a negative number. A negative exponent indicates a reciprocal. That is,

$$b^{-x} = (b^{-1})^x = \left( \frac{1}{b} \right)^x = \frac{1}{b^x}$$

All of the usual transformations of functions apply to exponential functions (below). One of the most important parts of an exponential function is the **base**. While any exponential phenomenon can be modeled using *any* base at all (by adjusting the rest of the transformation parameters to fit the data), we tend to stick to a small set of agreed-upon bases, like **2**, **10** and * e*. There will be much more to say about bases, especially

Remember that all of the transformations work in the same way on any function. Things are always more the same in algebra than different.

Move the sliders on this graph to get an idea of you two of the above transformations (**A** and **h**) work.

The top slider adjusts **A** between ±2 and the bottom one adjusts **h** between ±5. Notice that the **A** parameter increases the steepness of the rising function, and that if **A** is negative, the function is reflected across the x-axis.

As with all functions, adding or subtracting to/from the independent variable ( **x** ) translates the function to the left or right, respectively.

Knowing these transformations is important because we

- want to be able to sketch or mentally visualize an exponential function quickly, and
- want to be able to create a function that accurately models data and can make predictions about further measurements.

Consider the growth of money deposited in a bank account earning some annual interest rate, **r%**. (We use the letter **r** rather than a number so that our result will be good for *any* number.) One of the first challenges in solving problems with exponential functions is learning the new language. "**Annual interest**" means that every year, what's in the bank account is multiplied by the interest rate, **r**, and that amount is added to what's already there (the principal).

For every year after the initial deposit, this means that you're calculating this year's interest based not only on the initial amount deposited, but on *last* year's interest as well. That's called **compound growth** or **compound interest**.

Try to work your way through the table below. The algebra is pretty straightforward - just finding finding common factors between terms.

If an initial amount, A_{o} is deposited in a bank account earning r% interest, and not touched afterward, interest is earned each year. But after the first year interest is also earned on previous interest – that's compound interest

Year | Amount | Condensed |
---|---|---|

0 | $A_o$ (initial investment) | $A_o \, (1 + r)^0$ |

1 | $A_o + A_o \, r = A_o \, (1 + r)$ | $A_o \,(1 + r)^1$ |

2 | $A_o \, (1 + r) + r·A_o \, (1 + r) = A_o \, (1 + r)(1 + r) = A_o \, (1 + r)^2$ | $A_o \, (1 + r)^2$ |

3 | $A_o \, (1 + r)^2 + r·A_o \, (1 + r)^2 = A_o \, (1 + r)^3$ | $A_o \, (1 + r)^3$ |

. . . |
. . . |
. . . |

n years |
$A_o \, (1 + r)^n$ (the pattern continues) | $A_o \, (1 + r)^n$ |

For a system that grows at a rate, **r**, added periodically over **t** periods, the amount **A(t)** at any time **t** is

**n** with **t** for __time__.

$100 is deposited in a savings account that pays 5% interest per year. If no additional money is added, how much money will be in the account at the end of the 10^{th} year?

**Solution**

Now it's just a matter of completing the calculation to find A(10), the amount in the account after t = 10 years.

$$ \begin{align} A(10) &= 100(1 + 0.05)^{10} \\[5pt] &= 100 (1.05)^{10} \\[5pt] &= \$ \, 162.89 \end{align}$$

That's it. Just remember to convert the percent interest to decimal form: 5% = 5/100^{th}s, or 0.05.

If an initial investment of \$1000 earned $185 in interest after 8 years, what was the average rate of interest over that time?

**Solution**

Now the variable for which we must solve is inside a binomial raised to the 8^{th} power:

$$ \begin{align} 1185 &= 1000 (1 + r)^8 \\[5pt] 1.185 &= (1 + r)^8 \end{align}$$

To isolate **r**, we have to get rid of the 8^{th} power, and we do that by raising both sides of the equation to the ⅛ power (remember the laws of exponents).

$$ \begin{align} (1.185)^{1/8} &= [(1 + r)^8]^{1/8} \\[5pt] (1.185)^{1/8} - 1 &= r \\[5pt] r &= 0.0214 \\[5pt] r &= 2.14\, % \end{align}$$

So an interest (growth) rate of 2.14% per year would do the trick.

At the right temperature (37˚C, the temperature of your gut) and with enough nutrients, *E. coli *populations will double every 20 minutes. Starting with an initial population of one bacterium, with time, t, in units of 20 min., How many bacteria will be present in 8 hours?

**Solution***E. coli*. Finally, we're going to let 1 unit of time equal 20 minutes in this problem. It's not necessary at all, but it does simplify the arithmetic.

Start with the simple growth equation (I've used P for "population.")

Now we can solve for the rate. Not surprisingly, using this time scale, it's 100% over 20 minutes – the population *doubles*.

$$ \begin{align} 2 &= 1 (1 + r)^1 \\[5pt] r &= 1 \end{align}$$

100% growth every 20 min.

Now we can use **r** to construct our growth formula:

$$P(t) = P_o \cdot 2^t$$

Let's say you own a pizza restaurant and you buy a new oven. Your business is now worth more and will therefore be taxed accordingly. But the government (you and me) recognizes that the value of the oven will decrease over time, and provides a way to compensate you for that. It's called depreciation. Such an oven might depreciate by 5% per year. Calculate the taxable value of an oven originally purchased for \$10,000 after 10 years with a 5% depreciation rate.

**Solution****A(t) = A _{o}(1 + r)^{t}**, we use the depreciation form:

We start by writing the depreciation equation and what we know:

Then it's just a matter of finishing the calculation:

$$ \begin{align} A(t) &= 10,000(1 - 0.05)^{10} \\[5pt] &= 10,000 (0.95)^{10} \\[5pt] &= \$ 5,987 \end{align}$$

After ten years the pizza oven is worth a little more than half of its original value.

1. |
$1800 is placed in a bank account that earns 4.5% interest each year. Calculate the value of that initial investment after 10 years, if no money is added or withdrawn. ## Solution |

2. |
A car initially purchased for $32,500 is now 6 years old. If the annual depreciation rate (negative growth rate) is 6.8% per year, how much is the car worth today? ## Solution |

3. |
A population is thought to grow at an annual rate of 2.3%. How many years will it take such a population to double in size? ## Solution |

Keeping with our bank-account example, what if, instead of adding the interest to the balance at the end of each year, we added it in chunks periodically throughout the year. In banking, that's called **compounding** the interest. The idea will extend far beyond banking, so it's worth studying.

For example, let's say we have a bank account that pays 4% annual interest, and we do what banks call "compounding quarterly." That is, we split the 4% up into four equal pieces (1%) and pay that 1% in interest every three months. This begs two questions: (1) is there an easy way to do it, and (2) is it to our advantage?

To develop a formula for compounding, we can start with the simple growth formula and just modify it:

For a system that grows at a rate, **r**, compounded **n** times in each of **t** periods, the amount **A(t)** at any time **t** is

*Notice that the compound growth formula reverts to the simple growth formula when n = 1*

Now, of what value is this to a bank customer (and of what value is compounding in the mathematics of exponential growth)? Take a look at the figure below to see how.

Let's say we deposit \$1000 into an account that pays 4% interest per year, and we wait 10 years. According to the simple interest formula, we'd end up with about \$1480. The graph shows that there is about a \$9 advantage in switching to compounding quarterly (4 times a year).

Notice that as we increase the frequency of compounding, the amount of money gained in compounding more often (the "marginal gain") is measurable, but less. The dashed line is an **asymptote**.

That asymptote represents the limit as compounding becomes **continuous**, and that is the essence of the next section.

Note that the horizontal axis of this graph has no real scale; the data are just spread evenly across it to make it easy to read.

1. |
\$1500 is placed in a bank account that earns 1.8% interest, compounded quarterly (every three months, 4 times/year). Calculate the value of that initial investment after 10 years, if no money is added to or withdrawn from the account. ## Solution |

2. |
Which earns more money in 8 years, \$10,000 deposited in an account bearing 4% annual interest, or \$10,000 deposited in an account bearing 3.9%, compounded monthly? ## Solution |

3. |
Show that the compound interest formula is just the simple growth formula when n = 1. ## Solution |

If you take a look at the compound interest graph above, you'll notice that there is a limit to the amount of money that can be made by increasing the number of compounding steps. The word "steps" here is the key.

With compound growth, we wait a while, then add interest, then wait a while more and add interest, and so on. It's a stair-step process. But what if we took steps so frequently that the process was more like a *ramp* than a *stairway*?

That would be continuous growth, and the problem we'd be trying to solve is this one:

It's a tricky process to go through, but the answer to this question turns out to be very elegant. This equation, in the limit where **n → ∞**, converts to

where **e** is a very special number, the base of all continuously-growing exponential functions, **e = 2.7182818 ...** It's a transcendental number, like π, and you'll probably use it a lot, so get used to it.

We'll use this new function for continuous growth to model many types of exponential growth situations, especially populations. When populations become large enough, and births and deaths occur more-or-less simultaneously, the growth is approximately continuous, and the **P(t) = P _{o}e^{rt}** is a good model.

If you want to know more about where e comes from, go here:

For a system that grows continuously at a rate, **r**, the amount **A(t)** at any time **t** is

The base $e$ is on any scientific calculator — it's that important. On this TI-84 calculator, the $e^x$ function is the second function of the [ln] button. ln is the **natural log** function, ln(x). It's the inverse of the **natural exponential function**, $e^x.$ These functions are usually paired on the same calculator button.

You can learn more about logarithmic functions (you really can't do that much with exponential functions without "logs") in the logarithms section.

If you ever need the value of $e,$ just key in "$e^1$."

Compare the amount of money in a bank account after ten years at an interest rate of 4% per year, when that interest is compounded annually and quarterly. Assume that \$10,000 is initially deposited and that no more deposits are made.

The first part of the problem is a simple growth problem, so we'll write that function and what label what we know:

Now it's easy to solve:

$$ \begin{align} A(t) &= 10,000(1 + 0.04)^{10} \\[5pt] &= 10,000 (1.04)^{10} \\[5pt] &= 10,000(1.4802) \\[5pt] &= \$ \, 14,802 \end{align}$$

The second part is a compound-growth problem with n = 4 (quarterly):

And solving that gives us:

$$ \begin{align} A(t) &= 10,000 \left( 1 + \frac{0.04}{4} \right)^{4 \cdot 10} \\[5pt] &= 10,000 (1 + 0.01)^{40} \\[5pt] &= 10,000(1.01)^{40} \\[5pt] &= 10,000(1.4888) \\[5pt] &= \$ \, 14,888 \color{#E90F89}{\leftarrow \; \text{\$86 more!}} \end{align}$$

The compound interest option gives us more money at the end of ten years. Cool.

Compare the interest earned after ten years in a savings account started with \$1000.00 (no additional deposits) if the 5% interest is compounded **quarterly** (4 times/year), **weekly** (52 times/year), **daily** and **continuously**.

Quarterly:

$$ \begin{align} A(t) &= 10,000 \left( 1 + \frac{0.05}{4} \right)^{4 \cdot 10} \\[5pt] &= 10,000 (1 + 0.0125)^{40} \\[5pt] &= 10,000(1.0125)^{40} \\[5pt] &= 10,000(1.6436) \\[5pt] &= \$ \, 16,436 \end{align}$$

Weekly:

$$ \begin{align} A(t) &= 10,000 \left( 1 + \frac{0.05}{52} \right)^{52 \cdot 10} \\[5pt] &= 10,000 (1 + 0.000962)^{520} \\[5pt] &= 10,000(1.000962)^{520} \\[5pt] &= 10,000(1.6483) \\[5pt] &= \$ \, 16,483 \end{align}$$

Daily:

$$ \begin{align} A(t) &= 10,000 \left( 1 + \frac{0.05}{365} \right)^{365 \cdot 10} \\[5pt] &= 10,000 (1 + 0.000137)^{3650} \\[5pt] &= 10,000(1.000137)^{3650} \\[5pt] &= 10,000(1.6487) \\[5pt] &= \$ \, 16,487 \end{align}$$

Continuously – here let's pause to write our function and plug in values more carefully:

and the result is:

$$ \begin{align} A(t) &= 10,000 e^{0.05 \cdot 10} \\[5pt] &= 10,000 \cdot 1.6487 \\[5pt] &= \$ \, 16,487 \end{align}$$

Notice that the final amounts increase, but by less and less as we add more compounding steps, and that the difference between daily compounding and continuous growth is less than a cent, but the continuous case is still just a little greater – check it out for yourself.

A population of seabirds has grown reliably at a rate of 4% per year. If the population today is 16,000 birds, what was it 5 years ago?

**Solution**

We begin by writing the continuous-growth function and sketching in things we know

Now let's rearrange the function before plugging in numbers (always a good idea):

$$A_o = \frac{A(t)}{e^{rt}}$$

Now plug in what we know

$$A_o = \frac{16,000}{e^{5(0.04)}}$$

and the result is

$$A_o = 13,100 \; \text{birds}$$

Continuous growth models like this are almost always used to model populations.

1. |
A population grows continuously at a rate of 3.1% per year. At that rate, a population of 1 million people will grow to what size in 20 years? ## Solution |

2. |
A culture of ## Solution |

3. |
Under the right conditions, a certain virus takes 18 hours to double in number. Calculate the continuous rate of growth, ## Solution |

In problems that involve exponential decay, it is conventional to use the half-life measure. Here's how it works. Think about a pile of Iodine-131 (^{131}I). This radioactive isotope decays spontaneously to ^{131}Xe by emission of a beta particle followed by some gamma radiation. It's not possible to tell when a single atom of ^{131}I will decay, but we know that on average for a large enough quantity, one half of the initial amount will have decayed to ^{131}Xe in 8.02 days, the **half-life** of ^{131}I. Every 8.02 days, half of what remained will have decayed and become ^{131}Xe.

It's not important that you know the chemistry or physics of these problems right now, just that at any given time, what's left will be cut in half (turn to something else) in about 8 days.

Look at the graph to see how it works. It's pretty clear that this is a upside-down exponential graph.

To derive a formula for finding the amount of a radioactive substance still left after some number of half lives, we start with the formula for simple growth:

$$A(t) = A_o (1 + r)^t$$

Now if 1/2 of the sample is removed every time unit, then **r = -0.5**, so we have

$$A(t) = A_o \left( 1 - \frac{1}{2} \right)^t$$

Then we can introduce the **half life**, which we'll call **k**, to put the exponent in units of number of half lifes:

$$A(t) = A_o \left( \frac{1}{2} \right)^{t/k}$$

It's that exponent that's usually a little tricky. Think of it this way: If the half life is 8 days and we wait for 16 days, so **t/k** = 16/8 = 2 half lifes. We divide by **k** to convert the exponent to half lifes.

For exponential decay expressed in half-lifes, the function is:

where **t** is the time and **k** is the half life in the same units.

If the half life of a substance is 100 days, how much of the initial sample will be present after one year?

**Solution**_{o}) here. That's OK because the answer, expressed as a fraction (percent) of the starting amount, would be the same for *any* starting amount. For convenience, let's say we started with 1 (gram, Kilogram, whatever). Plugging in **k** = 100 days, **t** = 365 days and **A _{o}** = 1, we have:

Now it's just a matter of plugging in and solving for A(365).

So after one year, only 7.9% of the isotope will be present. The rest will have decayed to other stuff.

Some of these videos involve using logarithms to solve exponential problems. Don't worry about those until you learn about "logs," but those are crucial problems to learn, so do come back.

In this example, we derive the formula for simple exponential growth, **A(t) = A _{o}(1 + r)^{t}**, from scratch by considering a bank account that pays r% interest per year. The trickiest thing about this definition is factoring things out of an expression. Remember that

*Minutes of your life: 3:22*

In this example, we compare the interest on $1000.00 deposited for ten years (no other deposits) at 8% annual interest, **A(t) = A _{o}(1 + r)^{t}** and at 8% annual interest compounded quarterly:

*Minutes of your life: 2:29*

Given two data points, **P(1) = 0.1 million** and **P(7) = 1.2 million**, can we find a continuous-growth model, **P(t) = P _{o}e^{rt}** ? This really amounts to solving a two-equations, two-unknowns problem with logs.

*Minutes of your life: 2:14*

When plants and animals are alive, they exchange carbon atoms with the environment continually. When they die, that stops and the radioactive ^{14}**C** (one in every trillion carbon atoms) begins to decay. That decay can be measured. Using the half-life formula, **A(t) = A _{o}(1/2)^{t/k}** and the half life of

*Minutes of your life: 3:14*

Nuclear power plants generate ^{239}**Pu**, which is both toxic and highly radioactive. The half life of ^{239}**Pu** is 24,000 years. In this example we'll calculate how long it takes for any sample of ^{239}**Pu** to decay to 10% of its original mass. The answer is shocking!

*Minutes of your life: 2:52*

This is a simple growth problem using **A(t) = A _{o}(1 + r)^{t}**, but this time we use

*Minutes of your life: 2:19*

Things lose value over time, and we need a way to estimate the value of an object with the passing of time. Think about buying a new car. After a couple of years, it's a used car. You couldn't possibly sell it for what you originally paid. We call that loss of value **depreciation**, and we model it with the simple growth formula, **A(t) = A _{o}(1 + r)^{t}**, but where the rate

*Minutes of your life: 1:58*

How do we calculate **log _{3}82** ? 82 isn't a nice power of 3 (like 81 is), so it's difficult. Most calculators have a way to plug in logs with bases that aren't 10 (common logs) or e (natural logs), but you don't really need those. There's an easier way to do it and it's called the

*Minutes of your life: 3:26*

In this video, we solve two logarithmic equations, **log(x) + log(x + 3) = log(7)** and **2 ln(x) = ln(12x - 27)**. We use the laws of logs to re-express each so that the solutions are easy to get with simple algebra.

*Minutes of your life: 4:15*

Here we solve two logarithmic equations where the key is to recognize a common base and reexpress everything in terms of that base: **log _{16} = 1/4** and

*Minutes of your life: 2:04*

The key to graphing exponential and logarithmic functions is remembering that they're inverses, and have mirror symmetry across the line y = x. Here we sketch the graph of **y = e ^{x}**, and its inverse,

*Minutes of your life: 2:48*

Sketch the graph of an exponential function that has been transformed a bit: **f(x) = 3 ^{x-2}**. Then calculate and sketch its inverse on the same axes.

*Minutes of your life: 2:55*

Here are some manipulations of log and exponential functions that might help you come to terms with this, the most useful (and also most confusing – at first) law of logs.

*Minutes of your life: 2:04*

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