Doubling Time The current world population is
about 7.3 billion. Under current conditions, the
population is growing exponentially, with a yearly
growth factor of 1.011. In parts b and c, round your
answers to the nearest year.
a. Find a formula that gives the world population N,
in billions, after t years.
b. How long will it take for the population to double?
c. How long after doubling will it take for the
population to double again? accurate answers, show  working, and break it down for me 

Let's break down the problem step by step.

### a. Find a formula for the world population \( N \)

The population growth can be modeled using the formula for exponential growth:

\[
N(t) = N_0 \cdot r^t
\]

Where:
- \( N(t) \) is the population after \( t \) years,
- \( N_0 \) is the initial population,
- \( r \) is the growth factor per year,
- \( t \) is the time in years.

Given:
- \( N_0 = 7.3 \) billion,
- \( r = 1.011 \).

Thus, the formula becomes:

\[
N(t) = 7.3 \cdot (1.011)^t
\]

### b. How long will it take for the population to double?

To find the time it takes for the population to double, we set \( N(t) = 2 \cdot N_0 \):

\[
2 \cdot 7.3 = 7.3 \cdot (1.011)^t
\]

This simplifies to:

\[
2 = (1.011)^t
\]

To solve for \( t \), we take the logarithm of both sides:

\[
\log(2) = t \cdot \log(1.011)
\]

Now, solving for \( t \):

\[
t = \frac{\log(2)}{\log(1.011)}
\]

Calculating the logarithms:

- \( \log(2) \approx 0.3010 \)
- \( \log(1.011) \approx 0.00478 \)

Now substitute the values:

\[
t = \frac{0.3010}{0.00478} \approx 62.94 \text{ years}
\]

Rounding to the nearest year:

\[
t \approx 63 \text{ years}
\]

### c. How long after doubling will it take for the population to double again?

After doubling, the new population becomes \( 14.6 \) billion. We need to find out how long it takes for this population to double again to \( 29.2 \) billion.

Setting up the equation:

\[
29.2 = 14.6 \cdot (1.011)^t
\]

Dividing both sides by \( 14.6 \):

\[
2 = (1.011)^t
\]

The calculation is the same as in part b:

\[
t = \frac{\log(2)}{\log(1.011)} \approx 63 \text{ years}
\]

### Summary of Results
- **Formula:** \( N(t) = 7.3 \cdot (1.011)^t \)
- **Time to double:** \( 63 \text{ years} \)
- **Time to double again:** \( 63 \text{ years} \)

Would you like to go deeper into any of these steps or do you have any questions? Here are some related questions you might find interesting:

1. What factors contribute to the annual growth factor of 1.011?
2. How would the formula change if the growth rate decreased?
3. What are the implications of exponential growth for resource management?
4. How can we model population growth with varying growth rates?
5. What historical events have significantly influenced population growth rates? 

**Tip:** Consider how changes in birth rates, death rates, and migration affect population growth in real-world scenarios.

Doubling Time The current world population is about 7.3 billion. Under current conditions, the population is growing exponentially, with a yearly growth factor of 1.011. In parts b and c, round your answers to the nearest year. a. Find a formula that gives the world population N, in billions, after t years. b. How long will it take for the population to double? c. How long after doubling will it take for the population to double again? accurate answers, show working, and break it down for me.

Explore the calculation of the world population growth based on a current population of 7.3 billion and a growth factor of 1.011. Learn to derive the formula and find the time required for the population to double, which is approximately 63 years, along with subsequent doubling times.

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