Answer :
To determine how long it will take for [tex]$2,000 to grow to $[/tex]22,000 with an annual interest rate of 7% compounded monthly, we will use the compound interest formula:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount of money after time [tex]\( t \)[/tex].
- [tex]\( P \)[/tex] is the principal amount (initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (decimal form).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested.
In this problem:
- [tex]\( A = 22000 \)[/tex] (the final desired amount)
- [tex]\( P = 2000 \)[/tex] (the initial investment)
- [tex]\( r = 0.07 \)[/tex] (annual interest rate of 7%)
- [tex]\( n = 12 \)[/tex] (compounded monthly)
First, we need to rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 22000 = 2000 \left(1 + \frac{0.07}{12}\right)^{12t} \][/tex]
Now, simplify the inside of the parentheses:
[tex]\[ 22000 = 2000 \left(1 + \frac{0.07}{12}\right)^{12t} \][/tex]
[tex]\[ 22000 = 2000 \left(1 + 0.0058333\right)^{12t} \][/tex]
[tex]\[ 22000 = 2000 \left(1.0058333\right)^{12t} \][/tex]
Next, divide both sides by 2000 to isolate the exponential term:
[tex]\[ \frac{22000}{2000} = \left(1.0058333\right)^{12t} \][/tex]
[tex]\[ 11 = \left(1.0058333\right)^{12t} \][/tex]
To solve for [tex]\( t \)[/tex], we will take the natural logarithm (ln) of both sides:
[tex]\[ \ln(11) = \ln\left( \left(1.0058333\right)^{12t} \right) \][/tex]
Using the properties of logarithms ([tex]\(\ln(a^b) = b \cdot \ln(a)\)[/tex]):
[tex]\[ \ln(11) = 12t \cdot \ln(1.0058333) \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(11)}{12 \cdot \ln(1.0058333)} \][/tex]
Calculate the natural logarithms:
[tex]\[ \ln(11) \approx 2.397895 \][/tex]
[tex]\[ \ln(1.0058333) \approx 0.00582008 \][/tex]
So:
[tex]\[ t = \frac{2.397895}{12 \cdot 0.00582008} \][/tex]
[tex]\[ t \approx \frac{2.397895}{0.06984096} \][/tex]
[tex]\[ t \approx 34.34 \][/tex]
Thus, it will take approximately 34.34 years for [tex]$2,000 to grow to $[/tex]22,000 if it is invested at an annual interest rate of 7%, compounded monthly.
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount of money after time [tex]\( t \)[/tex].
- [tex]\( P \)[/tex] is the principal amount (initial investment).
- [tex]\( r \)[/tex] is the annual interest rate (decimal form).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year.
- [tex]\( t \)[/tex] is the number of years the money is invested.
In this problem:
- [tex]\( A = 22000 \)[/tex] (the final desired amount)
- [tex]\( P = 2000 \)[/tex] (the initial investment)
- [tex]\( r = 0.07 \)[/tex] (annual interest rate of 7%)
- [tex]\( n = 12 \)[/tex] (compounded monthly)
First, we need to rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 22000 = 2000 \left(1 + \frac{0.07}{12}\right)^{12t} \][/tex]
Now, simplify the inside of the parentheses:
[tex]\[ 22000 = 2000 \left(1 + \frac{0.07}{12}\right)^{12t} \][/tex]
[tex]\[ 22000 = 2000 \left(1 + 0.0058333\right)^{12t} \][/tex]
[tex]\[ 22000 = 2000 \left(1.0058333\right)^{12t} \][/tex]
Next, divide both sides by 2000 to isolate the exponential term:
[tex]\[ \frac{22000}{2000} = \left(1.0058333\right)^{12t} \][/tex]
[tex]\[ 11 = \left(1.0058333\right)^{12t} \][/tex]
To solve for [tex]\( t \)[/tex], we will take the natural logarithm (ln) of both sides:
[tex]\[ \ln(11) = \ln\left( \left(1.0058333\right)^{12t} \right) \][/tex]
Using the properties of logarithms ([tex]\(\ln(a^b) = b \cdot \ln(a)\)[/tex]):
[tex]\[ \ln(11) = 12t \cdot \ln(1.0058333) \][/tex]
Now, solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(11)}{12 \cdot \ln(1.0058333)} \][/tex]
Calculate the natural logarithms:
[tex]\[ \ln(11) \approx 2.397895 \][/tex]
[tex]\[ \ln(1.0058333) \approx 0.00582008 \][/tex]
So:
[tex]\[ t = \frac{2.397895}{12 \cdot 0.00582008} \][/tex]
[tex]\[ t \approx \frac{2.397895}{0.06984096} \][/tex]
[tex]\[ t \approx 34.34 \][/tex]
Thus, it will take approximately 34.34 years for [tex]$2,000 to grow to $[/tex]22,000 if it is invested at an annual interest rate of 7%, compounded monthly.
