We are given an initial investment of K4,000 at an annual interest rate of 6%, compounded annually. (a) We need to calculate the amount in the account for the years 2025 to 2029, assuming the investment starts in 2025. (b) We need to write an exponential function in the form $y = a \times b^x$ to represent this investment.

1. Problem Description

We are given an initial investment of K4,000 at an annual interest rate of 6%, compounded annually.

(a) We need to calculate the amount in the account for the years 2025 to 2029, assuming the investment starts in

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5. (b) We need to write an exponential function in the form $y = a \times b^x$ to represent this investment.

2. Solution Steps

(a) The formula for compound interest is:

$A = P(1 + r)^n$

where:

$A$ is the amount of money accumulated after n years, including interest.

$P$ is the principal amount (the initial amount of money).

$r$ is the annual interest rate (as a decimal).

$n$ is the number of years the money is invested or borrowed for.

In this case, $P = 4000$ and $r = 0.06$.

For 2025 (n = 1, assuming the investment starts in 2025):

$A_{2025} = 4000(1 + 0.06)^1 = 4000(1.06) = 4240$

For 2026 (n = 2):

$A_{2026} = 4000(1 + 0.06)^2 = 4000(1.06)^2 = 4000(1.1236) = 4494.4$

For 2027 (n = 3):

$A_{2027} = 4000(1 + 0.06)^3 = 4000(1.06)^3 = 4000(1.191016) = 4764.064$

For 2028 (n = 4):

$A_{2028} = 4000(1 + 0.06)^4 = 4000(1.06)^4 = 4000(1.26247696) = 5049.90784$

For 2029 (n = 5):

$A_{2029} = 4000(1 + 0.06)^5 = 4000(1.06)^5 = 4000(1.3382255776) = 5352.9023104$

(b) We need to write an exponential function in the form $y = a \times b^x$.

In this context, $y$ is the amount of money after $x$ years, $a$ is the initial investment, and $b$ is the growth factor (1 + interest rate).

Therefore, $a = 4000$ and $b = 1 + 0.06 = 1.06$.

The exponential function is $y = 4000 \times (1.06)^x$.

3. Final Answer

(a) The amounts in the account for the years 2025 to 2029 are:

2025: K4240

2026: K4494.40

2027: K4764.06

2028: K5049.91

2029: K5352.90

(b) The exponential function for this investment is $y = 4000 \times (1.06)^x$.