To determine how long it will take for a [tex]$1000 deposit to grow to $[/tex]2000 with an annual interest rate of 12%, compounded annually, we will use the formula for compound interest:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
where:
- [tex]\( A \)[/tex] is the amount of money accumulated after n years, including interest.
- [tex]\( P \)[/tex] is the principal amount (the initial sum of money).
- [tex]\( r \)[/tex] is the annual interest rate (decimal).
- [tex]\( n \)[/tex] is the number of times that interest is compounded per year.
- [tex]\( t \)[/tex] is the time the money is invested for in years.
Given:
- [tex]\( P = 1000 \)[/tex]
- [tex]\( A = 2000 \)[/tex]
- [tex]\( r = 0.12 \)[/tex]
- [tex]\( n = 1 \)[/tex] (compounded annually)
We need to solve for [tex]\( t \)[/tex].
First, the compound interest formula can be rearranged to solve for [tex]\( t \)[/tex]:
[tex]\[ 2000 = 1000 \left(1 + \frac{0.12}{1}\right)^{1*t} \][/tex]
This simplifies to:
[tex]\[ 2000 = 1000 \left(1.12\right)^t \][/tex]
Next, we divide both sides by 1000 to isolate the exponential expression:
[tex]\[ 2 = 1.12^t \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[ \ln(2) = \ln(1.12^t) \][/tex]
Using the logarithm property [tex]\(\ln(a^b) = b\ln(a)\)[/tex], we can write:
[tex]\[ \ln(2) = t \cdot \ln(1.12) \][/tex]
Now, we solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln(2)}{\ln(1.12)} \][/tex]
Calculating the natural logarithms:
[tex]\[ \ln(2) \approx 0.693147 \][/tex]
[tex]\[ \ln(1.12) \approx 0.113329 \][/tex]
Thus:
[tex]\[ t = \frac{0.693147}{0.113329} \approx 6.116255 \][/tex]
Rounding this result to the nearest hundredths, we get:
[tex]\[ t \approx 6.12 \][/tex]
Therefore, it will take approximately 6.12 years for the [tex]$1000 investment to grow to $[/tex]2000 with an annual interest rate of 12% compounded annually.
