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Difficulty: EasyExponential Functions and Equations

A scientist studying a sample of a radioactive isotope determines that its mass decays exponentially. The mass of the isotope, in grams, is modeled by the function M(t)=a⋅btM(t) = a \cdot b^t, where tt is the time, in days, since the study began, and aa and bb are constants. The table below shows the mass of the isotope for selected values of tt.

tt (days)M(t)M(t) (grams)
0128
196
272

What is the value of bb?

Answer: 0.75

Answer

0.75
To find the constant bb in the exponential decay function M(t)=a⋅btM(t) = a \cdot b^t, we can use the given table values. For t=0t = 0, M(0)=a⋅b0=a=128M(0) = a \cdot b^0 = a = 128. For t=1t = 1, M(1)=a⋅b1=a⋅b=96M(1) = a \cdot b^1 = a \cdot b = 96. Substituting a=128a = 128 into the second equation gives 128b=96128b = 96. Dividing both sides by 128128 yields b=96128=0.75b = \frac{96}{128} = 0.75 (or 34\frac{3}{4}). We can verify this with t=2t = 2: M(2)=128⋅(0.75)2=128⋅0.5625=72M(2) = 128 \cdot (0.75)^2 = 128 \cdot 0.5625 = 72, which matches the table.

Step-by-Step Solution

1
Determine the initial value aa using the data point for t=0t = 0.
a=128a = 128
Substituting t=0t = 0 into M(t)=a⋅btM(t) = a \cdot b^t gives M(0)=a⋅b0=aM(0) = a \cdot b^0 = a. Since the table shows M(0)=128M(0) = 128, aa must equal 128128.
2
Use the data point for t=1t = 1 to write an equation for bb.
128b=96128b = 96
Substituting t=1t = 1 and a=128a = 128 into the function gives M(1)=128⋅b1=128bM(1) = 128 \cdot b^1 = 128b. The table shows M(1)=96M(1) = 96.
3
Solve the equation to find bb.
b=0.75b = 0.75
Dividing both sides of 128b=96128b = 96 by 128128 yields b=96128b = \frac{96}{128}, which simplifies to 0.750.75 (or 34\frac{3}{4}).

Key Concept

Finding the decay factor (base) of an exponential function from a table of values.
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Topics

Exponential Functions and Equations

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