To determine how long it will take for an initial investment to grow from $1000 to $1500 at an interest rate of 4% per year, we can use the formula for compound interest:
\[ A = P(1 + r)^t \]
where:
- \( A \) is the amount of money accumulated after n years, including interest.
- \( P \) is the principal amount (the initial investment).
- \( r \) is the annual interest rate (decimal).
- \( t \) is the number of years the money is invested or borrowed.
In this case:
- \( A = 1500 \)
- \( P = 1000 \)
- \( r = 0.04 \)
We need to solve for \( t \):
\[ 1500 = 1000(1 + 0.04)^t \]
This simplifies to:
\[ 1500 = 1000(1.04)^t \]
Dividing both sides by 1000:
\[ 1.5 = (1.04)^t \]
Next, to solve for \( t \), we take the logarithm of both sides:
\[ \log(1.5) = \log((1.04)^t) \]
Using the power rule of logarithms:
\[ \log(1.5) = t \cdot \log(1.04) \]
Now, we can solve for \( t \):
\[ t = \frac{\log(1.5)}{\log(1.04)} \]
Using a calculator to find the logarithms:
- \( \log(1.5) \approx 0.1761 \)
- \( \log(1.04) \approx 0.0170 \)
Now substitute these values in:
\[ t \approx \frac{0.1761}{0.0170} \approx 10.35 \]
Rounding to the nearest tenth, we find:
\[ t \approx 10.3 \text{ years} \]
Thus, it will take approximately 10.3 years for the investment to grow from $1000 to $1500 at an interest rate of 4% per year.