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The interest on a sum of money that compounds continuously can be calculated with the formula I=Pe^{rt} - P , where P is the amount invested (the principal), r is the interest rate, and t is the amount of time the money is invested. If you invest $1000 in a bank account that pays 2.5% interest compounded continuously and you leave the money in that account for 4 years, how much interest will you earn?

Guidance

There are many special numbers in mathematics:  \pi , zero, \sqrt{2} , among others. In this concept, we will introduce another special number that is known only by a letter, e . It is called the natural number (or base), or the Euler number , after its discoverer, Leonhard Euler.

From the previous concept, we learned that the formula for compound interest is A=P \left(1+ \frac{r}{n}\right)^{nt} . Let’s set P, r and t equal to one and see what happens, A= \left(1+ \frac{1}{n}\right)^n .

Investigation: Finding the values of \left(1+ \frac{1}{n}\right)^n as n gets larger

1. Copy the table below and fill in the blanks. Round each entry to the nearest 4 decimal places.

n 1 2 3 4 5 6 7 8
\left(1+ \frac{1}{n}\right)^n \left(1+ \frac{1}{1}\right)^1 = 2 \left(1+ \frac{1}{2}\right)^2=2.25

2. Does it seem like the numbers in the table are approaching a certain value? What do you think the number is?

3. Find \left(1+ \frac{1}{100}\right)^{100} and \left(1+ \frac{1}{1000}\right)^{1000} . Does this change your answer from #2?

4. Fill in the blanks: As n approaches ___________, ___________ approaches e \approx 2.718281828459 \ldots

We define e as the number that \left(1+ \frac{1}{n}\right)^n approaches as n \rightarrow \infty ( n approaches infinity). e is an irrational number with the first 12 decimal places above.

Example A

Graph y=e^x . Identify the asymptote, y -intercept, domain and range.

Solution: As you would expect, the graph of e^x will curve between 2^x and 3^x .

The asymptote is y=0 and the y -intercept is (0, 1) because anything to the zero power is one. The domain is all real numbers and the range is all positive real numbers; y>0 .

Example B

Simplify e^2 \cdot e^4 .

Solution: The bases are the same, so you can just add the exponents. The answer is e^6 .

Example C

Gianna opens a savings account with $1000 and it accrues interest continuously at a rate of 5%. What is the balance in the account after 6 years?

Solution: In the previous concept, the word problems dealt with interest that compounded monthly, quarterly, annually, etc. In this example, the interest compounds continuously. The equation changes slightly, from A=P \left(1+ \frac{r}{n}\right)^{nt} to A=Pe^{rt} , without n , because there is no longer any interval. Therefore, the equation for this problem is A=1000e^{0.05(6)} and the account will have $1349.86 in it. Compare this to daily accrued interest, which would be A=1000 \left(1+ \frac{0.05}{365}\right)^{365(6)}=1349.83 .

Intro Problem Revisit Plug the given values into the equation I=Pe^{rt} and solve for I.

I=Pe^{rt} - P\\I = 1000\cdot e^{0.025\cdot 4} - 1000\\I = 1000 \cdot e^{0.1} - 1000\\I = 1000 \cdot 1.1052 - 1000\\I = 1105.20 - 1000 = 105.20

Therefore, at the end of 4 years, you will have earned $105.20 in interest.

Guided Practice

1. Determine if the following functions are exponential growth, decay, or neither.

a) y=\frac{1}{2}e^x

b) y=-4e^x

c) y=e^{-x}

d) y=2 \left(\frac{1}{e}\right)^{-x}

2. Simplify the following expressions with e .

a) 2e^{-3} \cdot e^2

b) \frac{4e^6}{16e^2}

3. The rate of radioactive decay of radium is modeled by R=Pe^{-0.00043t} , where R is the amount (in grams) of radium present after t years and P is the initial amount (also in grams). If there is 698.9 grams of radium present after 5,000 years, what was the initial amount?

Answers

1. Recall to be exponential growth, the base must be greater than one. To be exponential decay, the base must be between zero and one.

a) Exponential growth; e>1

b) Neither; a<0

c) Exponential decay; e^{-x}= \left(\frac{1}{e}\right)^x and 0< \frac{1}{e}<1

d) Exponential growth; \left(\frac{1}{e}\right)^{-x}=e^x

2. a) 2e^{-3} \cdot e^2 = 2e^{-1} or \frac{2}{e}

b) \frac{4e^6}{16e^2} = \frac{e^4}{4}

3. Use the formula given in the problem and solve for what you don’t know.

R &= Pe^{-0.00043t} \\698.9 &= Pe^{-0.00043(5000)} \\698.9 &= P(0.11648) \\6000 &= P

There was about 6000 grams of radium to start with.

Vocabulary

Natural Number (Euler Number)
The number e , such that as n \rightarrow \infty, \left(1+ \frac{1}{n}\right)^n \rightarrow e . e \approx 2.71828 .

Practice

Determine if the following functions are exponential growth, decay or neither. Give a reason for your answer.

  1. y=\frac{4}{3} e^x
  2. y=-e^{-x}+3
  3. y= \left(\frac{1}{e}\right)^x +2
  4. y= \left(\frac{3}{e}\right)^{-x} -5

Simplify the following expressions with e .

  1. e^{-3} \cdot e^{12}
  2. \frac{5e^{-4}}{e^3}
  3. 6e^5e^{-4}
  4. \left(\frac{4e^4}{3e^{-2}e^3}\right)^{-2}

Solve the following word problems.

The population of Springfield is growing exponentially. The growth can be modeled by the function P=Ie^{0.055t} , where P represents the projected population, I represents the current population of 100,000 in 2012 and t represents the number of years after 2012.

  1. To the nearest person, what will the population be in 2022?
  2. In what year will the population double in size if this growth rate continues?

The value of Steve’s car decreases in value according to the exponential decay function: V=Pe^{-0.12t} , where V is the current value of the vehicle, t is the number of years Steve has owned the car and P is the purchase price of the car, $25,000.

  1. To the nearest dollar, what will the value of Steve’s car be in 2 years?
  2. To the nearest dollar, what will the value be in 10 years?

Naya invests $7500 in an account which accrues interest continuously at a rate of 4.5%.

  1. Write an exponential growth function to model the value of her investment after t years.
  2. How much interest does Naya earn in the first six months to the nearest dollar?
  3. How much money, to the nearest dollar, is in the account after 8 years?

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Mar 12, 2013

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Aug 21, 2014

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