The number of milligrams [tex]\( D(h) \)[/tex] of a certain drug that is in a patient's bloodstream [tex]\( h \)[/tex] hours after the drug is injected is given by the following function:

[tex]\[ D(h) = 20 e^{-0.45 h} \][/tex]

When the number of milligrams reaches 9, 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.

[tex]\(\square\)[/tex] hours



Answer :

To determine the amount of time needed between injections, we need to solve for [tex]\( h \)[/tex] in the equation [tex]\( D(h) = 9 \)[/tex] where the given function is [tex]\( D(h) = 20 e^{-0.45h} \)[/tex].

Let's follow these steps to find [tex]\( h \)[/tex]:

1. Set [tex]\( D(h) \)[/tex] equal to 9:
[tex]\[ 20 e^{-0.45h} = 9 \][/tex]

2. Divide both sides by 20 to isolate the exponential term:
[tex]\[ e^{-0.45h} = \frac{9}{20} \][/tex]

3. Next, take the natural logarithm (ln) of both sides to eliminate the exponential expression:
[tex]\[ \ln\left(e^{-0.45h}\right) = \ln\left(\frac{9}{20}\right) \][/tex]

4. Utilize the property of logarithms [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[ -0.45h = \ln\left(\frac{9}{20}\right) \][/tex]

5. Solve for [tex]\( h \)[/tex] by dividing both sides by [tex]\(-0.45\)[/tex]:
[tex]\[ h = \frac{\ln\left(\frac{9}{20}\right)}{-0.45} \][/tex]

6. Calculate [tex]\( \ln\left(\frac{9}{20}\right) \)[/tex] and then divide by [tex]\(-0.45\)[/tex]:
[tex]\[ \ln\left(\frac{9}{20}\right) \approx \ln(0.45) \approx -0.7985076962177716 \][/tex]
[tex]\[ h = \frac{-0.7985076962177716}{-0.45} \approx 1.7744615471506036 \][/tex]

7. Finally, round the result to the nearest tenth:
[tex]\[ h \approx 1.8 \][/tex]

So, the amount of time needed between injections is approximately [tex]\( \boxed{1.8} \)[/tex] hours.