To solve this problem, we need to determine the initial investment amount [tex]\( P \)[/tex] that will grow to at least £4000 in 4 years with an annual compound interest rate of 3%.
The formula for compound interest is:
[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]
Where:
- [tex]\( A \)[/tex] is the final amount (£4000),
- [tex]\( P \)[/tex] is the principal amount (initial investment),
- [tex]\( r \)[/tex] is the annual interest rate (0.03 for 3%),
- [tex]\( n \)[/tex] is the number of times interest is compounded per year (1 in this case, as interest is compounded annually),
- [tex]\( t \)[/tex] is the time period in years (4 years).
We need to find [tex]\( P \)[/tex]. Rearranging the formula to solve for [tex]\( P \)[/tex], we get:
[tex]\[ P = \frac{A}{\left(1 + \frac{r}{n}\right)^{nt}} \][/tex]
Simplifying further for annual compounding ([tex]\( n = 1 \)[/tex]):
[tex]\[ P = \frac{A}{\left(1 + r\right)^t} \][/tex]
Now let's plug in the given values:
- [tex]\( A = 4000 \)[/tex]
- [tex]\( r = 0.03 \)[/tex]
- [tex]\( t = 4 \)[/tex]
Therefore:
[tex]\[ P = \frac{4000}{\left(1 + 0.03\right)^4} \][/tex]
[tex]\[ P = \frac{4000}{\left(1.03\right)^4} \][/tex]
Next, we calculate [tex]\(\left(1.03\right)^4\)[/tex]:
[tex]\[ \left(1.03\right)^4 = 1.03 \times 1.03 \times 1.03 \times 1.03 \][/tex]
[tex]\[ 1.03^2 = 1.0609 \][/tex]
[tex]\[ 1.0609 \times 1.03 = 1.092727 \][/tex]
[tex]\[ 1.092727 \times 1.03 = 1.12550881 \][/tex]
Round off the intermediate value if needed, but for accuracy, we use:
[tex]\[ 1.03^4 \approx 1.12550881 \][/tex]
Now, substitute this back into our formula to find [tex]\( P \)[/tex]:
[tex]\[ P = \frac{4000}{1.12550881} \][/tex]
[tex]\[ P \approx 3554.72 \][/tex]
We need to round this to the nearest pound:
[tex]\[ P \approx 3555 \][/tex]
Therefore, the smallest amount Louise can invest to ensure her investment grows to over £4000 at the end of 4 years is [tex]\(\boxed{£3555}\)[/tex].
