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To determine how many years it will take for a loan of [tex]$17,000 to grow to $[/tex]61,000 or more with an interest rate of 9% compounded annually, we can use the formula for compound interest:
[tex]\[ A = P(1 + r)^n \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount due after [tex]\( n \)[/tex] years.
- [tex]\( P \)[/tex] is the principal amount (initial loan).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( n \)[/tex] is the number of years.
Given:
- [tex]\( P = 17000 \)[/tex] dollars
- [tex]\( r = 0.09 \)[/tex] (since 9% is 0.09 when expressed as a decimal)
- [tex]\( A = 61000 \)[/tex] dollars
We need to find the smallest whole number [tex]\( n \)[/tex] such that:
[tex]\[ 17000(1 + 0.09)^n \geq 61000 \][/tex]
We will increment [tex]\( n \)[/tex] starting from 0 and evaluate the expression until it meets or exceeds 61,000.
By carrying out the calculations step-by-step for each increment in [tex]\( n \)[/tex]:
1. When [tex]\( n = 0 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^0 = 17000 \][/tex]
2. When [tex]\( n = 1 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^1 = 18530 \][/tex]
3. When [tex]\( n = 2 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^2 = 20297.7 \][/tex]
...
Continuing this process until we reach [tex]\( n = 15 \)[/tex]:
When [tex]\( n = 15 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^{15} \approx 61922.2 \][/tex]
At this point, the amount due is approximately [tex]\( 61922.2 \)[/tex], which is more than [tex]$61,000. Therefore, the smallest whole number of years needed for the loan to grow to at least $[/tex]61,000 is:
[tex]\[ \boxed{15} \][/tex]
[tex]\[ A = P(1 + r)^n \][/tex]
Where:
- [tex]\( A \)[/tex] is the amount due after [tex]\( n \)[/tex] years.
- [tex]\( P \)[/tex] is the principal amount (initial loan).
- [tex]\( r \)[/tex] is the annual interest rate (expressed as a decimal).
- [tex]\( n \)[/tex] is the number of years.
Given:
- [tex]\( P = 17000 \)[/tex] dollars
- [tex]\( r = 0.09 \)[/tex] (since 9% is 0.09 when expressed as a decimal)
- [tex]\( A = 61000 \)[/tex] dollars
We need to find the smallest whole number [tex]\( n \)[/tex] such that:
[tex]\[ 17000(1 + 0.09)^n \geq 61000 \][/tex]
We will increment [tex]\( n \)[/tex] starting from 0 and evaluate the expression until it meets or exceeds 61,000.
By carrying out the calculations step-by-step for each increment in [tex]\( n \)[/tex]:
1. When [tex]\( n = 0 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^0 = 17000 \][/tex]
2. When [tex]\( n = 1 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^1 = 18530 \][/tex]
3. When [tex]\( n = 2 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^2 = 20297.7 \][/tex]
...
Continuing this process until we reach [tex]\( n = 15 \)[/tex]:
When [tex]\( n = 15 \)[/tex]:
[tex]\[ 17000(1 + 0.09)^{15} \approx 61922.2 \][/tex]
At this point, the amount due is approximately [tex]\( 61922.2 \)[/tex], which is more than [tex]$61,000. Therefore, the smallest whole number of years needed for the loan to grow to at least $[/tex]61,000 is:
[tex]\[ \boxed{15} \][/tex]
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