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Precalculus: An Investigation of Functions, by David Lippman and Melonie Rasmussen, licensed under CC BY-SA 4.0. Read the original free This shelf’s study material is shared under the same license (CC BY-SA 4.0).
Chapter 8 · pp. 259–346
This chapter studies quantities that change by a fixed percent instead of a fixed amount. It starts with exponential functions, compound interest and continuous growth. It then moves to their graphs and to logarithms, the inverse functions that let us solve for an exponent. The last sections cover log properties, log graphs, real-world models (half-life, doubling time, cooling, earthquakes, pH) and fitting exponentials to data.
In linear growth, the output goes up by the same amount for every step in the input. A percent rate of change is different, because the amount added depends on how big the quantity already is. (p. 259)
Compare two companies that each start with 100 stores:
A adds 50 stores each year. B adds 50, then 75, then 112.5, because each year's gain is 50% of a larger total. (pp. 259–260)
There is a shortcut for B. Adding 50% of a number is the same as multiplying it by . After years, B has stores. (p. 260)
An exponential function has a constant percent rate of change. It can be written in either of two equivalent forms:
We only allow . Powers of negative numbers flip between positive and negative values, so they do not describe smooth growth or decay. For example, a 1.34% yearly growth gives , and a 13% yearly decay gives . (pp. 261, 263)
Modeling with a percent rate. Set the starting time to , take as the starting value, write as a decimal, and then evaluate at the time you need. India's population was 1.14 billion in 2008 and grew 1.34% a year, so
For 2020, use : billion. (p. 261)
How to spot an exponential situation. Ask whether the change is a fixed percent or a fixed amount. A 3.1% yearly raise is exponential, because each raise is a percent of the current salary. A $300 raise every year is linear, because the same amount is added each time. (p. 262)
If an annual rate is compounded monthly, divide the rate by 12 and count time in months. A 1.2% annual rate gives per month, so
After 24 months, the 1024.28. (p. 262)
Exponential decay is the case where is negative. Then lies between 0 and 1. Bismuth-210 loses about 13% each day, so and . The factor 0.87 is the 87% that remains each day. After 7 days, about 37.73 mg of the original 100 mg is left. (p. 263)
Reading an equation. To interpret , where is in thousands of contracts and is in quarters:
Starting value plus one more point. If you measure time from the first data point, you get for free. The deer population went from 80 in 2009 to 180 in 2015. With in years after 2009, the points are and . Then , so and . That is about 14.47% growth per year. (pp. 263–264)
Do not over-round. Early rounding distorts later predictions, especially for large inputs. Keep at least three significant digits in the growth rate. A factor like should be rounded no further than . (p. 264)
Two points, initial value unknown. Substitute both points into and solve the two equations together. For and :
The first equation gives . Putting that into the second gives , so and . Then . (pp. 265–266)
Reading from a graph. Any two clearly readable points are enough. From and we get and . Dividing the second by the first gives . Only the positive root is allowed, so . Then , and . (p. 266)
In compound interest, earnings are reinvested, so interest is paid on the accumulated balance and not only on the original amount. If the annual rate (APR, or nominal rate) is in decimal form, and interest is added times a year, each period earns . Over years there are periods, so
Here is the principal and is in years. (p. 267)
Example. Invest k=4$) for 10 years:
(p. 267)
Investing now to hit a target. Put the target in for and solve for . To have $40,000 after 18 years at 6% compounded twice a year:
Dividing gives a\approx\13{,}801$. (p. 268)
Compounding more often raises the balance, but each step up adds less. For $1000 at 10% for one year:
| Compounding | Balance after 1 year |
|---|---|
| Annually | $1100 |
| Semiannually | $1102.50 |
| Quarterly | $1103.81 |
| Monthly | $1104.71 |
| Daily | $1105.16 |
Moving from annual to semiannual adds 0.45. (pp. 268–269)
Because of compounding, the real yearly increase is more than the APR. The annual percentage yield measures it:
This is the value of $1 after one year, minus the original dollar. At 10% compounded daily, the APY is about 10.516%, which is above the 10% APR. (pp. 268–269)
Comparing savings offers. Work out the APY of each offer and choose the larger one. Bank A pays 1.2% compounded quarterly, so its APY is about 1.2054%. Bank B pays 1.1% compounded monthly, so its APY is about 1.1056%. Bank A is the better deal, because more frequent compounding at a lower rate does not make up the gap. (p. 269)
Imagine 2. Monthly it becomes about 2.714567, and every second it is about e\approx2.718282\left(1+\frac1k\right)^kke^x$ key for it (p. 270).
Because appears when compounding happens at tiny intervals, it gives us the continuous growth model:
Here is the starting amount and is the continuous rate. A negative means decay. This form suits smoothly changing things such as chemical reactions, large populations and radioactive decay (p. 270).
Example. Radon-222 decays at a continuous 17.3% per day. Starting with 100 mg, after 3 days we have mg (p. 271).
The continuous rate behaves like a nominal rate (APR), because it describes growth before compounding. The rate in behaves like an APY, because it is the real change over one period. So the two forms use different numbers for the same growth. For f(t)=1000e^{0.10t}1105.17, a true increase of 10.517%. The same account can be written (p. 271).
Compare and . Both are positive for every . The function is exponential growth: it climbs faster as increases and approaches 0 as decreases. The function is exponential decay. Since , we get , so each graph is the reflection of the other across the -axis (p. 277).
For with :
(pp. 277–278)
The long run behavior depends on . If , then as and as . If , the reverse happens. For , we have . The graph starts at , passes through , and shrinks toward 0 as grows (p. 278).
The parameters change the graph in different ways. Changing stretches the graph vertically and moves the intercept, but the long run behavior stays the same unless is negative, which flips the graph. A value of closer to 1 gives a less steep graph. For example, and both start at , but the second rises faster (pp. 279–280).
Most transformations can be absorbed into and :
Only a vertical shift gives something new. Every transformed exponential can be written
where is the horizontal asymptote and the vertical intercept is . Shifting down 4 gives , with asymptote (p. 281).
Example. Take . The factor flips the graph over and stretches it. The shift raises the asymptote to . The domain is all reals and the range is . As , , and as , . Curves like this model learning or growth that approaches a limit (p. 282).
To find a formula from a graph, first read the asymptote . Then use two other points. With asymptote and points and , the first point gives , so . The second gives , so and . Thus (p. 283).
Suppose 50 flies double every week, so . When do they reach 500? Dividing by 50 gives . Since and , is between 3 and 4, and a graph suggests about 3.3. Our earlier algebra cannot solve this exactly. Exponentials are one-to-one, so they have inverses, and the inverse is the logarithm (p. 287).
The logarithm base , written , is the inverse of . Because they undo each other:
For example, and (p. 287).
The key conversion is that is equivalent to . The log gives the exponent. So becomes , and becomes (pp. 288–289). This solves basic equations. If , then (p. 289).
Calculators evaluate only two bases. The common log has base 10, so and . The natural log has base . For example, (pp. 289–290).
The exponent property says . It moves an exponent to the front, and it works in either direction. Examples: , and (pp. 290–291).
The change of base formula lets us evaluate any base with a calculator:
So . The flies take about 3.32 weeks to grow from 50 to 500 (pp. 291–292).
Use these steps (pp. 292–294):
Example. For , divide by 5 to get . Taking logs gives , so
On a calculator, put parentheses around the whole denominator (p. 294).
Population target. India's population (in billions) was modeled by , with in years after 2008. Set it equal to 2 and divide by 1.14. Taking logs gives . So the population reaches 2 billion around 2050 (p. 293).
Acidity. The pH is , where is the hydrogen ion concentration in moles per liter. If , then . To go backward, a pH of 7 means , so moles per liter (p. 294).
To find a continuous rate from data, use the starting value for , substitute the second point, divide, and take . Suppose a population grows from 100 to 130 in 2 weeks. Then , so . Taking gives , a continuous rate of 13.12% per week (p. 295).
The two growth forms are linked. Setting gives
Here is the actual percent change per time unit and is the continuous (nominal) rate (pp. 295–296). For , . The annual growth is 12.75%, and the function can also be written (p. 296).
Earlier properties (inverse, exponent, change of base) cannot solve every log problem. Two more rules handle products and quotients inside a log:
The log of a product is the sum of the logs. The log of a quotient is the difference of the logs. Both rules need the same base on every log. (pp. 299–300)
Why the sum rule works. A log gives an exponent, so let and . Then and . Multiplying gives , because exponents add when you multiply powers of the same base. Now take of both sides. The inverse property gives . The difference rule is proved the same way, except you divide and the exponents subtract. (pp. 299–300)
Example (combining). Write as one log. The sum rule on the first two terms gives . The difference rule then gives . (p. 300)
Example (evaluating). Evaluate without a calculator.
Many students assume . This is false, because the log is not a linear function. It is a product that becomes a sum: .
A quick test: , but . The two sides are different, so the "rule" fails. (p. 299)
To break one log into several, work from the outside in. Use the quotient rule first, then the product rule, then the exponent rule.
Expand :
Read the same steps backward to combine logs. For example, , since . (p. 302)
Before calculators, people used the log rules to make hard arithmetic easier. Multiplying became adding. Dividing became subtracting. Powers and roots became multiplying and dividing. They looked up logs in printed tables or used slide rules. (pp. 302–303)
To find with a table, look up and . Their sum is . Searching the table backward shows this is , so .
For a cube root, . So . (pp. 302–303)
The rules also show how a change in input affects a log output. Since , let the original concentration be and the original pH be . If the concentration doubles to , then
Doubling the hydrogen ion concentration lowers the pH by . Multiplying the input of a log adds a constant to the output. (p. 303)
One log. Solve . There are two ways to start.
Either way, . Add 6 to get , then divide by 2 to get . Check: the inside is , so the answer is valid. (pp. 303–304)
Several logs on one side. First combine them into a single log, then rewrite in exponential form. Solve :
Check in the original equation. Both and are positive, so works. (p. 304)
Logs on both sides. Combine each side into a single log. Then exponentiate both sides, which removes the logs, and solve the algebra. Solve :
An extraneous solution comes out of the algebra but lies outside the domain of the original equation. You must discard it. A log can only take a positive input.
In the last example, would put in the original equation, and that is undefined. So is extraneous. Check : and , so is the only solution. Always substitute each answer back into the original equation, not the simplified one. (p. 305)
When you solve a log equation, the algebra can produce an answer that does not belong. An extraneous solution is an answer that falls outside the domain of the original equation, so it must be thrown out. In the textbook's Example 7, the algebra gave and . Putting back into the original equation leads to , which is undefined. So only works. Always check log-equation answers in the original equation. (p. 305)
Now consider an exponential equation with the variable in the exponent on both sides: . This models the populations of two countries (Kenya and Sudan) growing at different rates. We want the time when they are equal.
Method 1: take the log first.
The populations match after about 16 years. (p. 305)
Method 2: do algebra first.
Divide to gather the exponential parts on one side. Then use exponent rules to combine them into one power:
Now take one log of both sides, use the exponent property, and divide:
This looks different from Method 1, but the difference property of logs turns each log of a quotient into a difference of logs. That gives exactly the Method 1 expression. (pp. 305–306)
Both methods are correct. Isolating the exponential terms first and then taking a single log usually takes fewer steps than expanding everything with log properties. (p. 306)
Example (tanks). Tank A starts with 10 liters and loses 35% each week. Tank B starts with 30 liters and loses 50% each week. So and . Setting them equal and dividing gives , or . Then weeks. (p. 307)
The function is the inverse of . An inverse swaps inputs and outputs. For the exponential, . For the log, . The table for has from to and outputs . The log table uses those outputs as inputs and gives back to . (p. 310)
From the swap, we learn the following:
A quick way to sketch a log with base is to plot and . A larger base makes the graph grow more slowly. The common log, for example, needs an input of 100,000,000 just to reach an output of 8. (p. 311)
A log only accepts positive inputs. So set the expression inside the log greater than zero and solve.
Example. For , require . Subtract 5 to get . Dividing by flips the inequality, so . The domain is . (p. 311)
Some log transformations turn out to be the same as others.
Vertical stretch equals change of base. The change of base property gives
The factor is a constant that multiplies . So a vertical stretch or compression is the same as changing the base. For this reason, log graphs are usually written with the common or natural log. (p. 312)
Horizontal stretch or compression equals vertical shift. The sum property gives , and is just a constant. For example, , so the shifted graph can also be seen as a horizontal compression. A horizontal reflection is different. It cannot be rewritten as a vertical shift. (p. 312)
Horizontal shift is unique. No log property rewrites as another transformation. The graph moves left 2 units, the vertical asymptote moves to , and the domain becomes . (p. 313)
To sketch , factor the inside so you can read the transformations:
This is the common log with a horizontal reflection, a vertical reflection and stretch by 5, and a shift right 2. The asymptote is , and the domain is . Plot a couple of points: and . (pp. 313–314)
Any transformed log can be written as
or, if it is horizontally reflected,
The vertical asymptote is . Here is the vertical stretch (which acts like a change of base), is the vertical shift, and is the horizontal shift. (p. 314)
Steps:
Example. The graph has asymptote and is vertically reflected, so start with . The point gives , so . The point gives , so . The equation is
The second form comes from the change of base formula. (pp. 314–315)
The half-life of a radioactive isotope is the time needed for half of the substance to decay. Start with and ask when half remains: . The starting amount cancels, leaving . So half-life depends only on the decay factor, never on how much you start with. (p. 318)
Finding half-life from a decay rate. Bismuth-210 loses about 13% each day, so . Set and take a log of both sides. The exponent property moves down:
The half-life is about 5 days. (pp. 318–319)
Repeated halving. Each time one half-life passes, the amount is cut in half again. Cesium-137 has a half-life of about 30 years. Starting with 200 mg, 100 mg is left after 30 years, 50 mg after 60 years and 25 mg after 90 years. (p. 319)
Finding an annual rate from a half-life. Use and the fact that half remains at . Then . Take the 30th root and subtract 1:
Cesium-137 decays about 2.284% per year. (p. 319)
Carbon-14 dating. Scientists usually use the continuous form . Carbon-14 has a half-life of 5730 years, so . Taking the natural log gives . A bone with 20% of its original carbon-14 satisfies . Solving gives years. (p. 320)
The doubling time of a growing quantity is the time it takes to double. For , set . Again cancels and you get , so the starting amount does not matter. (p. 321)
From two data points. A cancerous growth had 300 cells last month and 360 this month. Let be this month, so the points are and . Then , and gives . The model is . Solve :
(pp. 321–322)
From a known doubling time. Suppose a website doubles its users every 5 months. In the continuous model (thousands of users), doubling in 5 months means . So . After 12 months, thousand users. In general, doubling time gives . (p. 322)
A hot object in cooler air loses heat quickly at first and then levels off toward the air temperature. That leveling off is a horizontal asymptote. So the model is a decay curve shifted vertically:
Here is the surrounding temperature, is the starting temperature minus , and is negative for cooling. (p. 323)
Example. A cheesecake leaves the oven at 165°F and goes into a 35°F refrigerator. Then , so . After 10 minutes it is 150°F, so . This gives . To reach 70°F, solve . Isolate the exponential first: . Then minutes. (pp. 323–324)
When values range over huge sizes, a normal number line bunches the small ones together. Planet distances from the sun run from 58 to 4500 million km. Their logs run only from about 1.76 to 3.65, so they spread out evenly. A logarithmic scale places the logarithms of the values at equal spacing. (pp. 325–326)
To read a point on a log scale, convert back with a power of 10. A point halfway between and has , so . To place 6000, compute and mark that position. (p. 326)
Equal distances mean equal ratios. On a log scale, moving one tick multiplies the value by 10. Compare $1,000,000 with $10,000, and $1000 with $10. Both pairs have a ratio of 100, so both gaps look the same. (p. 327)
One order of magnitude means about ten times larger. The number of orders of magnitude by which exceeds is . For and , the ratio is and , so there are 3 orders of magnitude. (p. 327)
The Moment Magnitude Scale is often wrongly called the Richter scale. It gives
where is the seismic moment, a measure of earth movement. To go from magnitude to , solve for it. For , , so . (p. 328)
Magnitude 8 versus 6. Magnitude 8 gives . The ratio is . So a rise of 2 in magnitude means 1000 times more earth movement. (pp. 328–329)
Doubling the movement. If doubles, then
The constant is added to the old magnitude. A magnitude 3.0 quake with twice the movement is about magnitude 3.2. This shows the main feature of any log scale: multiplying the quantity adds to the scale value. (pp. 329–330)
A semi-log graph has one axis on a linear scale and the other on a log scale. A log-log graph has both axes logarithmic. (p. 339) Stock charts often use a log vertical axis because it shows percentage change. The Dow fell from about 14,000 to 8,000 in 2008, about 43%. It fell from 380 to 42 from 1929 to 1932, about 89%. The dollar drop was smaller, but the percentage drop was much larger. (pp. 338–339)
Why exponentials look straight. For , the sum and exponent properties give
This is a line in . Its vertical intercept is and its slope is . For , the logs are 0.477 at and 1.079 at , and these lie on a line. (pp. 339–340)
Going backward. Suppose a semi-log line passes through and . The slope is , so . Rewrite this as a power of 10:
(pp. 340–341)
Linearization. To fit an exponential when a tool only does linear regression:
(p. 341) Applied to import data, this gave , which predicts about 722 billion dollars in 2011. (p. 343)
A caution. Hard drive costs fit , with in years after 1980. This gives , a drop of about 41% per year. But the model predicts $622.3 per MB in 1980, while the actual cost was $192.31. The fit matches only in order of magnitude, so be careful when predicting from it. (pp. 342–343)
