math word problem

Question

The number of milligrams D(h) of a certain drug that is in a patient's bloodstream h hours after the drug is injected is given by the following function.

D(h)=25e^−0.35h

When the number of milligrams reaches 6, the drug is to be injected again. How much time is needed between injections? Round your answer to the nearest tenth, and do not round any intermediate computations. 

Answer 1

Hello,

To solve an exponential function we will be utilizing the properties of logarithms.

we have the function:

D(h)=25e^−0.35h

The question prompts us to solve for h (the numbers of hours) that would have passed when the D(h) (the amount of medication in the patient's bloodstream) equals 6mg in order to know when the patient needs to be injected again.

Our equation will look like this:

6 = 25e^−0.35h

to solve we will start by dividing both sides by 25

6/25 = (25e^−0.35h)/25

6/25 = e^−0.35h

Next, I will substitute x for -0.35h for the meantime, and 6/25 = .24 with no need to round so I will also simplify there

So the equation will look like this:

.24 = e^x

Recall the properties of logarithms Log_by=x is equivalent to b^x = y

and log

Using this property we can rewrite our equation

b = e

x = -0.35h

y = .24

We now have:

Log_e.24 = -0.35h

Recall the properties of natural logs:

Log_ey is equivalent to ln(y)

We can apply that property and it will look like the following:

Log_e.24 = -0.35h

=> ln(.24) = -.35h

=> -1.427116356 = -.35h

=> -1.427116356/-0.35 = -0.35h / -0.35

=> 4.0774... = h

=> h = 4.1

So after 4.1 hours, the patient needs to be injected again.