Math 30P Exponents Logarithms Sequences Series Regression/Recursive Interest Growth Richter/PH Prerequisite Skills
exponential function: a function of the form y = Ab^{x}
step function: a function whose graph increases in discrete amounts and is constant between increments
exponential growth: growth that can be modelled by an exponential function
The number of bacteria, N(t), in a a culture is given by the formula , where t is the elapsed time (in minutes) and d is the doubling time. How many bacteria were in the initial account. Solution Method 1 The formula for doubling time is , N_{o} is the initial account. According to the equation given, N_{o} = 300 Method 2 300 bacteria were in the initial count. 
Example 2: Doubling Time Initially a bacteria culture contains 1250 bacteria. The count increased to approximately 80 000 bacteria 1 1/2 hours later. What is the doubling period for this bacterium? Solution Let N(t) represent the number of bacteria after t hours; let N_{0} represent the initial bacteria count; and let d represent the doubling time. The doubing period for this bacterium is 0.25 h or 15 min. 
Example 2: Doubling Time A single bacterium has a mass of 1.0 × 10^{15} kg. If its doubling time is 30 min and if there are sufficient nutrients and the wastes they produced do not inhibit growth, what would the total mass of the bacteria be after 3 days? How would that mass compare to the mass of Earth, which is approximately 6.0 × 1024 kg? SolutionLet y be the number of bacteria at any time, t (in hours). Since the initial population is 1 bacterium and the number is doubling every 30 min (or 0.5 h), the function that models the population after t hours is Convert days into hours. 3d × 24 h/d = 72 h Now, find the number of bacteria after 3 days. Multiply the number of bacteria by the mass of 1 bacterium. The total mass of the bacteria would be about 2.23 × 1028 kg. Compare the mass of the bacteria to the mass of the Earth. The mass of the bacteria would be approximately 3700 time the mass of the Earth. In the preceding example you investigated exponential growth in terms of its doubling time. For bacteria, the time required for the population to double is also called the generation time. 
Example 4: Doubling Time The initial bacteria culture doubles in 3 hours. Write an equation that represents the time elapsed for the culture? Solution 
What to Learn 
Help to Remember What to Learn


The exponential function N(t) = N_{o}(r)^{t/h}, N_{o} > 0, 0 < r < 1, h > 0, where N_{o}, r, t and h are real numbers.

Interactive Activity


Solving Exponential Decay Equations 
The halflife of strontium90 is 25 years. A 100 g sample of strontium90 is purchased. How much strontium90 will remain in 60 years. In 60 years there will be 19.95 g remaining in the strontium90 sample. 
Example 2 The halflife of carbon14 is 5730 years. This is the time it takes for 50% of the original carbon14 atoms to decay to nitrogen. The percentage, P(t), of the original carbon14 that remains after t years can be modelled by the following exponential function: If the remaining percentage is 25%, then the age, t (in years), of the artifact can be found by solving the following exponential equation: When there is 25% of the carbon14 remaining, the artifact will be 11 460 years old. 
Example 3 The halflife of of iodine131 is 8.1 days. How long will it take for 100 g of iodine131 to decay to 77.4 g. Solution The general equation is , where A(t) represents the amount present after t days, A_{0} is the original amount, and H is the halflife of the substance. A(t) = 77.4 g, A_{0} = 100 g, and H = 8.1 days. It will take approximately 3.0 days for 100 g of iodine131 to decay to 77.4 g. 
Example 4 The halflife of strontium90 is 25 years. How much time has elapsed if 1/4 remains in a sample. Solution Let A(t) represent the amount of strontium90 after t years, and let A_{0} represent the original amount. Since the substance is decreasing by onehalf every 25 years, the equation is If ¼ remains after t years, then Therefore, 50 years have elapsed. 
Example 5 What is the half life of plutonium243 if 1/64 of the original amount remains after 30 h? Solution Let A(t) represent the amount of plutonium243 after t hours; let A_{0} represent the original amount; and let h be the halflife. Since the substance is decreasing by onehalf every h hours, the equation is If 1/64 of the original amount remains after 30 h, then Therefore, the halflife of plutonium243 is 5 h. 
Example 6 The half life of cobalt60 is 5.3 years. After 10.6 years what fraction of the cobalt60 will remain? How long will it take until 12.5% of the original cobalt60 remains. Solution After 10.6 years, t = 10.6. After 10.6 years, only ¼ of the original amount remains. After 15.9 years, only 12.5% of the original amount remains. 
Parts of this work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education
Math 30P Exponents Logarithms Sequences Series Regression/Recursive Interest Growth Richter/PH Prerequisite Skills