Algebra Help: More Exponential Word Problems

In an earlier article I introduced the Algebra student to exponential equations, exponential functions, and exponential word problems. This material is covered in Saxon Algebra 2, Lesson 115. Saxon, however, because of the limits of his math text book, only really discusses one form of exponential word problem in the text; it was this style of exponential word problem I examined in my first article. To provide the math student with a broader understanding of the material I will be discussing two variations of exponential word problems the Algebra student should be able to solve.

General Form of an Exponential Equation
All exponential functions (or exponential equations) can be represented using the following format:

A_t = A₀*e^k*t

Where A_t is the amount at time t, A₀ is the amount at time 0 (initial amount) and k is some constant specific to the exponential equation. e is the special irrational number called the "natural logarithm"; this is approximately equal to 2.718.

Sample Exponential Word Problem #1
In the beginning there were 100 tons of radioactive waste. It decayed exponentially. In one year there was only 95.5 tons of radioactive waste left. In how many years (from the beginning) will there be 50 tons of radioactive waste left?

Since we know that this is an exponential word problem, we always start with our general equation:

A_t = A₀*e^k*t

From the math word problem we can pull out the following information:

A₀ = 100
A₁ = 100*e^k*1 = 95.5

Our goal is to solve for k, our constant, so that we can calculate the specific equation for our exponential function.

Divide both sides by 100 and rearrange our equation:

e^k = 0.955

To solve for k we need to rewrite 0.955 in terms of base e. For that we use our natural logarithm, ln function on our calculater. ln 0.955 = -0.046.

e^k = e^-0.046, so k = -0.046

We now know our equation can be written as:

A_t = 100*e^-0.046*t

Our math word problem asks us to find when the amount is equal to 50 tons. We have to set our equation equal to 50, then solve for t, our time variable.

50 = 100*e^-0.046*t

We use the same steps we used to solve for k, divide, then use our natural logarithm function ln:

0.5 = e^-0.046*t
ln 0.5 = -0.693

e^-0.693 = e^-0.046*t
-0.693 = -0.046*t
t = 15.1 years (approximately)

Sample Exponential Word Problem #2
For his biology lab Seth carefully recorded how many bacteria were in the petri dish on day 2 (225 bacteria) and on day 4 (750 bacteria), but he forgot to record how many bacteria he started with. Using these two data points, estimate how many bacteria he started with (How many bacteria were there in the beginning?)

Since we know that this is an exponential word problem, we always start with our general equation:

A_t = A₀*e^k*t

From the math word problem we can pull out the following information:

A₂ = A₀*e^k*2 = 225
A₄ = A₀*e^k*4 = 750

We need to find both k and A₀.

Solve the first equation for A₀ = 225 /(e^k*2)

Substitute into the second equation:

(225 /(e^k*2))*e^k*4 = 750

Rewrite the equation so it is easier to read and understand:

225*e^-2*k*e^4*k = 750
225*e^2*k = 750

Divide both sides by 225, then use our natural logarithm ln:

e^2*k = 3.33
e^2*k = e^1.20
2*k = 1.20
k = 0.60

We can now input k into our equation for A₀:

A₀ = 225 /(e^k*2) = 225/(e^0.60*2) = 67.77

Seth started with approximately 68 bacteria in his petri dish.

Source
John H. Saxon, Jr. Algebra 2: An Incremental Development. Third Edition