To find the time required for an investment of $2500 to amount to $3000 at an interest rate of 6% compounded quarterly, we can use the formula for compound interest:
\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]
where:
- \( A \) is the future amount of money (final amount),
- \( P \) is the principal amount (initial investment),
- \( r \) is the annual interest rate (as a decimal),
- \( n \) is the number of times that interest is compounded per year,
- \( t \) is the time in years.
Given:
- \( A = 3000 \)
- \( P = 2500 \)
- \( r = 0.06 \) (which is 6% as a decimal)
- \( n = 4 \) (since the interest is compounded quarterly)
We substitute these values into the formula:
\[ 3000 = 2500 \left(1 + \frac{0.06}{4}\right)^{4t} \]
First, simplify \( \frac{0.06}{4} \):
\[ \frac{0.06}{4} = 0.015 \]
Now the equation becomes:
\[ 3000 = 2500 \left(1 + 0.015\right)^{4t} \]
This simplifies to:
\[ 3000 = 2500 \left(1.015\right)^{4t} \]
Next, we divide both sides by 2500:
\[ \frac{3000}{2500} = \left(1.015\right)^{4t} \]
Calculating \( \frac{3000}{2500} \):
\[ \frac{3000}{2500} = 1.2 \]
Now we have:
\[ 1.2 = \left(1.015\right)^{4t} \]
To solve for \( t \), we take the natural logarithm of both sides:
\[ \ln(1.2) = \ln\left((1.015)^{4t}\right) \]
Using the property of logarithms \( \ln(a^b) = b \cdot \ln(a) \), we can rewrite the right side:
\[ \ln(1.2) = 4t \cdot \ln(1.015) \]
Now, solve for \( t \):
\[ t = \frac{\ln(1.2)}{4 \ln(1.015)} \]
Now we compute the values:
- Calculate \( \ln(1.2) \):
\[ \ln(1.2) \approx 0.182321 \]
- Calculate \( \ln(1.015) \):
\[ \ln(1.015) \approx 0.014888 \]
Substituting these values back in gives:
\[ t = \frac{0.182321}{4 \cdot 0.014888} \approx \frac{0.182321}{0.059552} \approx 3.063 \]
Thus, the time required for $2500 to amount to $3000 at 6% compounded quarterly is approximately:
\[ \boxed{3.06} \text{ years} \]