Infant mortalitly rate is reducing 8.5% per year. How long will it take for it to be reduced by 36%

The infant mortality rate is reducing by 8.5% per year, and we want to find how long it will take for it to be reduced by 36%.

This can be modeled as an exponential decay problem. The formula for exponential decay is:

\[
P(t) = P_0 \times (1 - r)^t
\]

Where:
- \( P(t) \) is the remaining percentage after time \( t \),
- \( P_0 \) is the initial percentage (100%, since we are reducing from the full rate),
- \( r \) is the rate of decay (8.5% per year, or 0.085),
- \( t \) is the time in years.

We need to find the time \( t \) when the reduction is 36%, meaning the remaining percentage is 64%. So:

\[
0.64 = 1 \times (1 - 0.085)^t
\]

Now we solve for \( t \).

\[
0.64 = (0.915)^t
\]

Taking the natural logarithm on both sides:

\[
\ln(0.64) = t \times \ln(0.915)
\]

Solving for \( t \):

\[
t = \frac{\ln(0.64)}{\ln(0.915)}
\]

Let's calculate the value of \( t \).It will take approximately 5 years for the infant mortality rate to be reduced by 36%.

Would you like more details or have any other questions?

Here are 5 related questions:
1. How does an 8.5% reduction per year compare to a constant rate of reduction?
2. What if the rate of reduction changed over time? How would that affect the result?
3. How would the result differ if the reduction was compounded monthly instead of annually?
4. What would the time be if the target reduction were 50% instead of 36%?
5. Can this model be applied to population growth or other exponential processes?

**Tip:** Exponential decay problems often use logarithms to solve for time, so understanding logarithmic properties is crucial in these types of equations.

Infant mortality rate is reducing 8.5% per year. How long will it take for it to be reduced by 36%?

Discover how to calculate the time required for an 8.5% annual reduction in infant mortality to result in a 36% decrease. This problem uses the exponential decay formula and logarithmic calculations, making it suitable for students in Grades 10-12.

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