Question
Clayton is looking up census data for his state at the local library. His state conducts a census every decade. He reads that the population was 441,000 the year he was born, and that it had increased to 449,820 a decade later. The librarian told Clayton that the population of the state is expected to continue increasing each decade.
Write an exponential equation in the form y=a(b)x that can model the population in Clayton's state, y, x decades after he was born.
Use whole numbers, decimals, or simplified fractions for the values of a and b.
y =
How many decades after Clayton was born will the census find the population to be greater than 500,000?
decades
Write an exponential equation in the form y=a(b)x that can model the population in Clayton's state, y, x decades after he was born.
Use whole numbers, decimals, or simplified fractions for the values of a and b.
y =
How many decades after Clayton was born will the census find the population to be greater than 500,000?
decades
Answers
Answered by
GPT-4o mini
To create an exponential equation of the form \( y = a(b)^x \), we need to determine the values of \( a \) and \( b \) based on the census data provided.
1. When Clayton was born (let's call this \( x = 0 \)), the population was \( y = 441,000 \). Thus, we have:
\[
a = 441,000
\]
2. A decade later (when \( x = 1 \)), the population increased to \( y = 449,820 \). We can use this information to solve for \( b \):
\[
449,820 = 441,000(b)^1
\]
Dividing both sides by 441,000 gives:
\[
b = \frac{449,820}{441,000} \approx 1.02
\]
Now we can write the exponential equation:
\[
y = 441,000(1.02)^x
\]
Next, we want to find how many decades after Clayton was born the population will exceed 500,000:
\[
500,000 < 441,000(1.02)^x
\]
To solve for \( x \):
1. Divide both sides by 441,000:
\[
\frac{500,000}{441,000} < (1.02)^x
\]
This simplifies to:
\[
1.134 < (1.02)^x
\]
2. Now take the logarithm of both sides:
\[
\log(1.134) < x \log(1.02)
\]
3. Solve for \( x \):
\[
x > \frac{\log(1.134)}{\log(1.02)}
\]
Calculating this using a calculator:
- \( \log(1.134) \approx 0.053 \)
- \( \log(1.02) \approx 0.009 \)
Thus,
\[
x > \frac{0.053}{0.009} \approx 5.89
\]
Since \( x \) must be a whole number (representing complete decades), we round up to the next whole number:
\[
x = 6
\]
### Final Summary:
The exponential equation modeling the population is:
\[
y = 441,000(1.02)^x
\]
The census will find the population to be greater than 500,000 **6 decades** after Clayton was born.
1. When Clayton was born (let's call this \( x = 0 \)), the population was \( y = 441,000 \). Thus, we have:
\[
a = 441,000
\]
2. A decade later (when \( x = 1 \)), the population increased to \( y = 449,820 \). We can use this information to solve for \( b \):
\[
449,820 = 441,000(b)^1
\]
Dividing both sides by 441,000 gives:
\[
b = \frac{449,820}{441,000} \approx 1.02
\]
Now we can write the exponential equation:
\[
y = 441,000(1.02)^x
\]
Next, we want to find how many decades after Clayton was born the population will exceed 500,000:
\[
500,000 < 441,000(1.02)^x
\]
To solve for \( x \):
1. Divide both sides by 441,000:
\[
\frac{500,000}{441,000} < (1.02)^x
\]
This simplifies to:
\[
1.134 < (1.02)^x
\]
2. Now take the logarithm of both sides:
\[
\log(1.134) < x \log(1.02)
\]
3. Solve for \( x \):
\[
x > \frac{\log(1.134)}{\log(1.02)}
\]
Calculating this using a calculator:
- \( \log(1.134) \approx 0.053 \)
- \( \log(1.02) \approx 0.009 \)
Thus,
\[
x > \frac{0.053}{0.009} \approx 5.89
\]
Since \( x \) must be a whole number (representing complete decades), we round up to the next whole number:
\[
x = 6
\]
### Final Summary:
The exponential equation modeling the population is:
\[
y = 441,000(1.02)^x
\]
The census will find the population to be greater than 500,000 **6 decades** after Clayton was born.