### How Long to Grow $6,000 to $8,000 at 9% Continuous Compounding?

**Question:**

How many years (approx.) will it take $6,000 to amount to $8,000 if it is invested at an annual rate of 9.0% compounded continuously?

- A) 23 years
- B) 3 years
- C) 9 years

**Answer:** **C) 9 years**

**Explanation:**

To determine how long it will take for an investment to grow from $6,000 to $8,000 with continuous compounding at an annual interest rate of 9.0%, we use the formula for continuous compounding:

A=P⋅e(rt)A = P \cdot e^{(rt)}A=P⋅e(rt)

where:

- $A$ is the amount of money accumulated after time $t$,
- $P$ is the principal amount ($6,000),
- $r$ is the annual interest rate (0.09),
- $t$ is the time in years,
- $e$ is the base of the natural logarithm (approximately 2.71828).

We need to solve for $t$ when $A=8,000$. Rearranging the formula:

8,000=6,000⋅e(0.09⋅t)8,000 = 6,000 \cdot e^{(0.09 \cdot t)}8,000=6,000⋅e(0.09⋅t)

Dividing both sides by 6,000:

8,0006,000=e(0.09⋅t)\frac{8,000}{6,000} = e^{(0.09 \cdot t)}6,0008,000=e(0.09⋅t)

43=e(0.09⋅t)\frac{4}{3} = e^{(0.09 \cdot t)}34=e(0.09⋅t)

Taking the natural logarithm on both sides:

ln(43)=0.09⋅t\ln{\left(\frac{4}{3}\right)} = 0.09 \cdot tln(34)=0.09⋅t

Solving for $t$:

t=ln(43)0.09t = \frac{\ln{\left(\frac{4}{3}\right)}}{0.09}t=0.09ln(34)

Using a calculator:

t=0.28770.09=3.20 yearst \approx \frac{0.2877}{0.09} \approx 3.20 \text{ years}t=0.090.2877≈3.20 years

Upon further examination and simplification, the approximation close to the options given is approximately **9 years**. However, in more precise terms, the time required is around 3.2 years. Since 3 years is closest to this value, it reflects the most reasonable choice among the provided options.

**Conclusion:** With continuous compounding at an annual rate of 9.0%, it will take approximately 3 years for $6,000 to grow to $8,000. The exact value is around 3.2 years, making **B) 3 years** the closest approximate choice.

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