We are given an initial investment of K4,000 at an annual interest rate of 6%, compounded annually. We need to: (a) Calculate the amount in the account for the years 2025 to 2029. (b) Write an exponential function representing the investment in the form $y = a \times b^x$. (c) Calculate the amount in the account in the year 2050. (d) Determine the year when the account balance will reach K10,000.

1. Problem Description

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

(a) Calculate the amount in the account for the years 2025 to

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9. (b) Write an exponential function representing the investment in the form $y = a \times b^x$.

(c) Calculate the amount in the account in the year

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5

0. (d) Determine the year when the account balance will reach K10,

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2. Solution Steps

(a) Calculate the amount in the account for years 2025 to

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9. Let's assume the investment starts at the beginning of the year

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5. The formula for compound interest is:

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

where:

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

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

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

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

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

Year 2025: $t = 0$, so $A = 4000(1 + 0.06)^0 = 4000(1) = 4000$

Year 2026: $t = 1$, so $A = 4000(1 + 0.06)^1 = 4000(1.06) = 4240$

Year 2027: $t = 2$, so $A = 4000(1 + 0.06)^2 = 4000(1.06)^2 = 4000(1.1236) = 4494.40$

Year 2028: $t = 3$, so $A = 4000(1 + 0.06)^3 = 4000(1.06)^3 = 4000(1.191016) = 4764.064 \approx 4764.06$

Year 2029: $t = 4$, so $A = 4000(1 + 0.06)^4 = 4000(1.06)^4 = 4000(1.26247696) = 5049.90784 \approx 5049.91$

(b) Write an exponential function for this investment in the form $y = a \times b^x$.

Here, $y$ represents the amount after $x$ years, $a$ is the initial investment, and $b$ is $(1 + r)$.

So, $y = 4000 \times (1.06)^x$

(c) How much would be in the account in year 2050?

The investment starts in

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5. So, the number of years between 2025 and 2050 is $2050 - 2025 = 25$.

$A = 4000(1.06)^{25} = 4000(4.29187074795) \approx 17167.48$

(d) In what year would you expect to have K10,000 in the account?

We need to find $x$ such that $10000 = 4000(1.06)^x$.

Divide both sides by 4000:

$2.5 = (1.06)^x$

Take the natural logarithm of both sides:

$ln(2.5) = x \times ln(1.06)$

$x = \frac{ln(2.5)}{ln(1.06)} = \frac{0.91629073187}{0.05826890812} \approx 15.725$ years.

Since the investment starts in 2025, the year when the account balance reaches K10,000 is approximately $2025 + 15.725 = 2040.725$. Since we're looking for the year, we round up to the next year, so it will be in the year

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3. Final Answer

(a) Amount in the account:

2025: K4000

2026: K4240

2027: K4494.40

2028: K4764.06

2029: K5049.91

(b) Exponential function:

$y = 4000 \times (1.06)^x$

(c) Amount in 2050:

K17167.48

(d) Year to reach K10,000:

2041