In order to solve problems involving exponential decay, it is necessary to:

Exponential decay is generally applied to word problems that involve financial applications as well as those that deal with radioactive decay, medicine dosages, and

population decline. To decay exponentially means that the topic being studied is decreasing in

proportion to the amount that was previously present. The following is an example of an exponential decay problem.

When doctors prescribe medicine, they must consider how much the drug’s effectiveness will decrease as time passes. If each hour a drug is only 95% as effective as the previous hour, at some

point the patient will not be receiving enough medicine and must be given another dose. If the initial dose was 250 mg and the drug was administered 3 hours ago, how long will it take for the initial dose to reach a dangerously low level of 52 mg?

First, we will need to use the general exponential decay formula:

In the formula,

represents the amount of medicine after time has passed.

represents the initial amount of medicine. The constant

*a* represents the

rate of decay (and is always a number between 0 and 1), and

*t* stands for time, which is in hours in this problem.

Now, we need to substitute known values for the variables in the formula. The problem asks how long it will take the initial dose to become dangerously low. Therefore,

is 52 in this problem.

is the initial dose which is 250 mg. The

rate of decay

is

which will be converted to the decimal 0.95. Time

*t* is what we are trying to find. So we have the following:

Finally we must solve the

equation for time

*t*. To do so, first divide both sides by 250 to simplify the equation.

Next we take the log of each

side of the

equation and bring down the exponent,

*t*. For a reminder on taking the log of both sides as well as the properties of logs, please examine this

companion lesson.

Now, to solve for time *t*, divide both sides by (log 0**.**95) to obtain the following:

Now we use a calculator to find the value for *t*

*
*

hours

Checking our answer shows

A(t) = 250(0.95)

^{30.61} = 52.00673226

In fact, *t* actually represents less than the number of hours required for the amount of drug left to go below 52 mg. This is because there is still slightly more than 52 mg left at time t = 30.61 hours, our rounded off answer. The amount of the drug left will go below 52 mg sometime AFTER 30.61 hours has passed.