Print this Page

9.2 – Exponential Functions

Key Terms

  • Base – A number in exponential form. It appears beneath the exponent.
  • e – The irrational number e 2.7182818284590…
    • You can approximate e by substituting large values of n into the expression (1+\frac{1}{n})^{n}
  • Compound Interest – Interest applied to both the principal and any previously earned interest.
  • Exponent – A small raised number that tells you how many times to multiply the base number by itself.
  • Exponential Decay – A condition in which a quantity decreases exponentially over time.
    • The quantity decreases at a rate that is proportional to the current value of the quantity.
    • Exponential decay can be modeled by the equation A_{t}=A_{0}e^{-kt}, where A_{t} = the amount at time t, A_{0} = the initial amount, t = time, and k = a constant called the decay constant.
    • If a quantity decays exponentially, its half-life is a constant.
    • A situation in which a quantity decreases by a common ratio at regular intervals.
  • Exponential Function – A function that has the form F(x)=a\bullet b^{x}, where the coefficient a is a constant, the base b is positive but not equal to 1, and the exponent x is any number.
  • Exponential Growth – A condition in which a quantity increases exponentially over time.
    • The quantity increases at a rate that is proportional to the current value of the quantity.
    • Exponential growth can be modeled by the equation A_{t}=A_{0}e^{kt}, where A_{t} = the amount at time t, A_{0} = the initial amount, t = time, and k = a constant called the growth constant.
    • If a quantity grows exponentially, its doubling time is a constant.
    • A situation in which a quantity increases by a common ratio at regular intervals.
  • Half-Life – The time it takes for a quantity to decrease to one-half of its size or value.
    • If a quantity is decaying exponentially, its half-life is a constant.
  • Input – A number that is entered into a function. The input variable in a function is the independent variable.
  • Output – The result of a function. The output variable in a function is the dependent variable.
  • Principal – The amount of money invested or borrowed.

Notes

What if…
  • What if someone gave you 2 cents ($0.02) each day… how long would it take for you to reach $1,000,000?
    • 1,000,000 divided by 0.02 gives you 50,000,000 days!  That’s 137,000 years!
    • This is an example of linear growth.
  • However, what if someone gives you 2 cents on June 1st with one of two options (below).  Which would you rather:
    1. Would you prefer to double your pennies from the previous day for a month, or
    2. Would you rather receive 1 million dollars right now?

  •  The exponential function that represents this problem is F(x)=2^{x}.
    • x: day of the month
    • 2: doubling (base)
    • You start with zero pennies, but then someone gives you 2 pennies on day 1: F(1)=2^{1}=2.

Alg1B 6.2 Pennies2


  • So, would you rather have $1,000,000 dollars today, or double your pennies everyday for a month?
    • I think I’ll take the pennies!

 

  • Exponential Growth and Decay

Alg1B 6.2 Exp Growth Decay

Exponential Functions
  • Functions with a variable in the exponent
  • Formula for Exponential Functions (General Form)

Alg1B 6.2 Exp Fx

  • Examples of Exponential Functions
    • F(x)=2\bullet 3^{x}
    • F(x)=-4\bullet 5^{x}
    • F(x)=(\frac{1}{2})^{x}
    • F(x)=\frac{2}{3}\bullet 0.1^{x}

 

Inputs and Outputs with Exponential Functions
  • As the input values increase, the function with the variable exponent grows more quickly than the function with the variable base.
    • x: input
    • F(x): outputs

Alg2B 9.2 InputOutput

 

Exponential Growth Exponential Decay
  • An exponential growth function represents a quantity that has a constant doubling time.
  • Ex. -4\bullet 5^{2t}
  • Ex. 2\bullet 3^{6t}
  • An exponential decay function represents a quantity that has a constant halving time (half-life).
  • The exponent will either be negative (like 3^{-x}) or the base will be between 0 and 1 (like 0.3^{x}).
  • Ex. 2\bullet 0.4^{3t}
  • Ex. 8^{-2t}
  • Exponential growth and decay functions are written in standard form as F(t)=A_0\bullet b^{kt}, where A_0 is an initial amount, b is the growth factor, k is the growth rate, and t is time.
    • A_0, b, and k all stand for fixed numbers and t is the only variable.

Alg2B 9.2 GrowthDecay

 

Compounding Interest Periodically
  • Compound interest is an example of exponential growth because as more time passes, the amount of money grows faster and faster.
    • Interest is money owed (and paid) for borrowing money.
    • When you put your money in a bank account, it’s like loaning the money to the bank. For that service, the bank pays you a percentage of the amount you put in.
    • If you borrow money from the bank, you have to pay interest to the bank.
  • The rule of 72 is a quick way to estimate the effects of compound interest.
    • Rule: The number of years it takes an amount to double is approximately the interest rate divided by 72.
      • Years to double ≈ 72 ÷ interest rate
      • So at 3%, it would take about 24 years for a $200 investment to double to $400.
  • Formula for Compound Interest

Alg1B 6.2 Compound Interest Formula

  • A: Amount of money accumulated in the account. Amount depends on time: A depends on t, so A(t).
    t: Time that money remains invested at the given interest rate. Number of periods.
    P: Principle is the original amount invested.
    r: Interest Rate. Ex. 9% = 0.09.  (To write a percent as a decimal, divide the percent number by 100).
    n: Number of times that interest is compounded per year.

Alg1B 6.2 Compound Times

  • Example

Alg1B 6.2 Principal Rate Time

 

Compounding Interest Continuously
  • Continuous compounding means that the interest instead gets calculated at every instant in time, not just at periodic intervals.
  • So, what if you invest $1 at 100% interest for 1 year?
    • This is known as continuous compounding.

Alg1B 6.2 Compound Interest

Alg1B 6.2 Compound ContinuouslyAnswer: When the interest is compounded continuously,
the value of the account is e dollars (about $2.72)!


  • Formula for Compounding Interest Continuously
    • A(t)=P\bullet e^{rt}
      • A(t): amount of money in the account after t years
      • P: principal – original amount (in dollars) you put into the account
      • r: interest rate (written as a decimal)
      • t: number of years the interest is compounded
  • The number e
    • e\approx 2.7182818284590...
    • The number e is an irrational number.
    • This means that its decimal representation goes on and on forever and does not repeat.
    • You can round the value of e to 2.718, unless otherwise instructed.
    • You can approximate e by substituting large values of n into the expression: (1+\frac{1}{n})^{n}
    • To evaluate e in the Windows calculator, type the exponent first, then “Inv,” then e^{x}.
      • Ex. Evaluate e^{-2}.
      • Type 2, \pm, Inv, e^{x}
      • Answer: 0.135
  • Example: You invest $500 in a savings account. The account pays 4% interest compounded continuously. How much money is in the account after 7 years?
    • P: $500
    • r: 0.04
    • t: 7
    • Setup: A(7)=500\bullet e^{7\bullet 0.04}
      • Simplify: A(7)=500\bullet e^{0.28}
      • Simplify: A(7)=500\bullet 1.3231298
      • Answer: $661.55

Examples

  • Ex 1. Find F(3), when F(x)=3^{x}
    • Substitute: F(3)=3^{3}
    • Answer: 3\bullet 3\bullet 3=27
  • Ex 2. Find F(3), when F(x)=(\frac{1}{6})^{x}
    • Substitute: F(3)=(\frac{1}{6})^{3}
    • Answer: (\frac{1}{6})\bullet (\frac{1}{6})\bullet (\frac{1}{6})=(\frac{1}{216})
  • Ex 3. Find F(4), when F(x)=(\frac{1}{3})\bullet 4^{x}
    • Substitute: F(4)=(\frac{1}{3})\bullet 4^{4}
    • Answer: (\frac{1}{3})\bullet 4\bullet 4\bullet 4\bullet 4=\frac{256}{3}
  • Ex 4. Find F(3), when F(x)=4\bullet (\frac{1}{3})^{x}
    • Substitute: F(3)=4\bullet (\frac{1}{3})^{3}
    • Answer: 4\bullet (\frac{1}{3})\bullet (\frac{1}{3})\bullet (\frac{1}{3})
  • Ex 5. Find F(-4), when F(x)=2^{-4}
    • Substitute: F(-4)=\frac{1}{2^{4}}
    • Answer: (\frac{1}{2\bullet 2\bullet 2\bullet 2})=\frac{1}{16}
  • Ex 6. Find F(2), when F(x)=2\bullet (\frac{1}{2^{3t}})
    • Substitute: F(2)=2\bullet (\frac{1}{2^{3\bullet 2}})
    • Simplify: 2\bullet (\frac{1}{2^{6}})
    • Expand: 2\bullet (\frac{1}{2\bullet 2\bullet 2\bullet 2\bullet 2\bullet 2})
    • Answer: 2\bullet (\frac{1}{64})=\frac{1}{32}
  • Ex 7. How much would $200 invested at 6% interest compounded annually be worth after 6 years? Round your answer to the nearest cent.
    • Formula: A(t)=P(1+\frac{r}{n})^{nt}
    • Substitute: A(6)=200(1+\frac{0.06}{1})^{1\bullet 6}
    • Simplify: A(6)=200(1.06)^{6}
    • Simplify More: A(6)=200(1.06)^{6}
    • Answer: $283.70
  • Ex 8. How much would $500 invested at 6% interest compounded monthly be worth after 5 years? Round your answer to the nearest cent.
    • Formula: A(t)=P(1+\frac{r}{n})^{nt}
    • Substitute: A(5)=500(1+\frac{0.06}{12})^{12\bullet 5}
    • Simplify: A(5)=500(1+0.005)^{60}
    • Simplify More: A(5)=500(1.005)^{72}
    • Answer: $674.43
  • Ex 9. How much would $400 invested at 9% interest compounded continuously be worth after 3 years? Round your answer to the nearest cent.
    • Formula: A(t)=P\bullet e^{rt}
    • Substitute: A(3)=400\bullet e^{0.09\bullet 3}
    • Simplify: A(3)=400\bullet 2.718^{0.27}
    • Answer: $523.97

Permanent link to this article: http://newvillagegirlsacademy.org/math/?page_id=4621