Answer :
Given the exponential function \( f(x) = 555(1.026)^x \) models the population of a country in millions, where \( x \) is the number of years after 1971:
### Part (a)
Substitute \( x = 0 \):
\[ f(0) = 555(1.026)^0 = 555(1) = 555 \]
So, the country's population in 1971 was 555 million.
### Part (b)
Substitute \( x = 27 \) (since 1998 is 27 years after 1971):
\[ f(27) = 555(1.026)^{27} \]
Using a calculator to find \( 555 \times (1.026)^{27} \):
\[ f(27) \approx 555 \times 2 = 1110 \]
So, the country's population in 1998 was approximately 1110 million.
### Part (c)
Substitute \( x = 54 \) (since 2025 is 54 years after 1971):
\[ f(54) = 555(1.026)^{54} \]
Using a calculator to find \( 555 \times (1.026)^{54} \):
\[ f(54) \approx 555 \times 4 = 2220 \]
So, the country's population in 2025 is predicted to be approximately 2220 million.
### Part (a)
Substitute \( x = 0 \):
\[ f(0) = 555(1.026)^0 = 555(1) = 555 \]
So, the country's population in 1971 was 555 million.
### Part (b)
Substitute \( x = 27 \) (since 1998 is 27 years after 1971):
\[ f(27) = 555(1.026)^{27} \]
Using a calculator to find \( 555 \times (1.026)^{27} \):
\[ f(27) \approx 555 \times 2 = 1110 \]
So, the country's population in 1998 was approximately 1110 million.
### Part (c)
Substitute \( x = 54 \) (since 2025 is 54 years after 1971):
\[ f(54) = 555(1.026)^{54} \]
Using a calculator to find \( 555 \times (1.026)^{54} \):
\[ f(54) \approx 555 \times 4 = 2220 \]
So, the country's population in 2025 is predicted to be approximately 2220 million.