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Characteristics of the exponential function, y = A 0 e kt

An exponential function with the form y = A 0 e k t has the following characteristics:

  • one-to-one function
  • horizontal asymptote: y = 0
  • domain: ( ,   )
  • range: ( 0 , )
  • x intercept: none
  • y-intercept: ( 0 , A 0 )
  • increasing if k > 0 (see [link] )
  • decreasing if k < 0 (see [link] )
Two graphs of y=(A_0)(e^(kt)) with the asymptote at y=0. The first graph is of when k>0 and with the labeled points (1/k, (A_0)e), (0, A_0), and (-1/k, (A_0)/e). The second graph is of when k<0 and with the labeled points (-1/k, (A_0)e), (0, A_0), and (1/k, (A_0)/e).
An exponential function models exponential growth when k > 0 and exponential decay when k < 0.

Graphing exponential growth

A population of bacteria doubles every hour. If the culture started with 10 bacteria, graph the population as a function of time.

When an amount grows at a fixed percent per unit time, the growth is exponential. To find A 0 we use the fact that A 0 is the amount at time zero, so A 0 = 10. To find k , use the fact that after one hour ( t = 1 ) the population doubles from 10 to 20. The formula is derived as follows

  20 = 10 e k 1     2 = e k Divide by 10 ln 2 = k Take the natural logarithm

so k = ln ( 2 ) . Thus the equation we want to graph is y = 10 e ( ln 2 ) t = 10 ( e ln 2 ) t = 10 · 2 t . The graph is shown in [link] .

A graph starting at ten on the y-axis and rising rapidly to the right.
The graph of y = 10 e ( ln 2 ) t
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Half-life

We now turn to exponential decay . One of the common terms associated with exponential decay, as stated above, is half-life , the length of time it takes an exponentially decaying quantity to decrease to half its original amount. Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.

To find the half-life of a function describing exponential decay, solve the following equation:

1 2 A 0 = A o e k t

We find that the half-life depends only on the constant k and not on the starting quantity A 0 .

The formula is derived as follows

1 2 A 0 = A o e k t 1 2 = e k t Divide by  A 0 . ln ( 1 2 ) = k t Take the natural log . ln ( 2 ) = k t Apply laws of logarithms . ln ( 2 ) k = t Divide by  k .

Since t , the time, is positive, k must, as expected, be negative. This gives us the half-life formula

t = ln ( 2 ) k

Given the half-life, find the decay rate.

  1. Write A = A o e k t .
  2. Replace A by 1 2 A 0 and replace t by the given half-life.
  3. Solve to find k . Express k as an exact value (do not round).

Note: It is also possible to find the decay rate using k = ln ( 2 ) t .

Finding the function that describes radioactive decay

The half-life of carbon-14 is 5,730 years. Express the amount of carbon-14 remaining as a function of time, t .

This formula is derived as follows.

         A = A 0 e k t The continuous growth formula . 0.5 A 0 = A 0 e k 5730 Substitute the half-life for  t  and  0.5 A 0  for  f ( t ) .        0.5 = e 5730 k Divide by  A 0 . ln ( 0.5 ) = 5730 k Take the natural log of both sides .           k = ln ( 0.5 ) 5730 Divide by the coefficient of  k .          A = A 0 e ( ln ( 0.5 ) 5730 ) t Substitute for  r  in the continuous growth formula .

The function that describes this continuous decay is f ( t ) = A 0 e ( ln ( 0.5 ) 5730 ) t . We observe that the coefficient of t , ln ( 0.5 ) 5730 1.2097 × 10 −4 is negative, as expected in the case of exponential decay.

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The half-life of plutonium-244 is 80,000,000 years. Find function gives the amount of carbon-14 remaining as a function of time, measured in years.

f ( t ) = A 0 e 0.0000000087 t

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Radiocarbon dating

The formula for radioactive decay is important in radiocarbon dating , which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby, who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about 1% error for plants or animals that died within the last 60,000 years.

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
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-24m+3+3mÁ^2
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-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
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x=3-2y
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3x-12y=18
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
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The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
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I'm guessing, but it's somewhere around $4335.00 I think
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12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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