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The problem involves solving for \( h \) (time in hours) in the equation:

\[
D(h) = 35e^{-0.35h}
\]

where \( D(h) \) is the drug concentration. We're tasked to find \( h \) when \( D(h) = 7 \). The steps are as follows:

1. Substitute \( D(h) = 7 \) into the formula:
   \[
   7 = 35e^{-0.35h}
   \]

2. Divide through by 35 to isolate the exponential term:
   \[
   e^{-0.35h} = \frac{7}{35} = 0.2
   \]

3. Take the natural logarithm (ln) of both sides to remove the exponential:
   \[
   \ln(e^{-0.35h}) = \ln(0.2)
   \]

4. Simplify using the property \( \ln(e^x) = x \):
   \[
   -0.35h = \ln(0.2)
   \]

5. Solve for \( h \):
   \[
   h = \frac{\ln(0.2)}{-0.35}
   \]

Now, let me calculate this value. math

# Given values
numerator = math.log(0.2)  # ln(0.2)
denominator = -0.35        # Coefficient of h

# Solve for h
h = numerator / denominator
hThe time between injections is approximately \( h = 4.6 \) hours when rounded to the nearest tenth.

Would you like a detailed explanation of the steps or additional assistance? 

Here are some related questions to expand understanding:
1. How does the base of natural logarithms \( e \) influence the decay rate?
2. What happens if the decay constant (0.35) changes? How does it affect \( h \)?
3. How is the formula \( D(h) \) derived in pharmacokinetics?
4. Why is taking the natural logarithm effective in solving exponential equations?
5. Can this method be applied to model other drug concentrations over time?

**Tip:** When solving exponential equations, remember to isolate the exponential term before applying logarithms for simplicity.

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 function D(h) = 35e^(-0.35h). When the number of milligrams reaches 7, 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.

Learn how to calculate the time required between injections based on the exponential decay of drug concentration. This example uses the function D(h) = 35e^(-0.35h) to find when the concentration reaches 7 mg, covering concepts like logarithms and exponential equations. Suitable for Grades 10-12.

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