We need to find the number of years it takes for an investment of $800 to grow to $2000 at an annual interest rate of 6% compounded annually.

1. Problem Description

We need to find the number of years it takes for an investment of $800 to grow to$2000 at an annual interest rate of 6% compounded annually.

2. Solution Steps

We can use the compound interest formula:

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

where:

$A$ is the future value of the investment

$P$ is the principal amount (initial investment)

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

$t$ is the number of years

In this problem, we have:

$A = 2000$

$P = 800$

$r = 0.06$

We want to find $t$.

Substitute the given values into the formula:

$2000 = 800(1 + 0.06)^t$

$2000 = 800(1.06)^t$

Divide both sides by 800:

$\frac{2000}{800} = (1.06)^t$

$2.5 = (1.06)^t$

To solve for $t$, we can use logarithms. Take the natural logarithm of both sides:

$ln(2.5) = ln((1.06)^t)$

$ln(2.5) = t \cdot ln(1.06)$

Now, divide both sides by $ln(1.06)$:

$t = \frac{ln(2.5)}{ln(1.06)}$

$t \approx \frac{0.91629}{0.05827}$

$t \approx 15.725$

Since we are looking for the number of years, we round to the nearest whole number. In this case, the closest whole number is

1

6.

3. Final Answer

16