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.
