Reuben deposited $2000 into an account with a 3.8% annual interest rate, compounded semiannually. Assuming that no withdrawals are made, how long will it take for the investment to grow to $3000?

3000 = 2000 [1 + (.038 / 2)]^(2t)

... t is the number of years

1.5 = 1.019^(2t)

log(1.5) = 2t * log(1.019)

To determine how long it will take for the investment to grow to $3000, we need to use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A = future value
P = principal amount
r = annual interest rate
n = number of times interest is compounded per year
t = time in years

In this case, we have:
P = $2000
A = $3000
r = 3.8% or 0.038 (as a decimal)
n = 2 (since interest is compounded semiannually)

Substituting these values into the formula, we get:
$3000 = $2000(1 + 0.038/2)^(2t)

To solve for t, we need to isolate it on one side of the equation. We can start by dividing both sides by $2000:

$3000/$2000 = (1 + 0.038/2)^(2t)

After simplifying, we have:
1.5 = (1.019)^2t

Next, take the logarithm of both sides of the equation to solve for t. We can use the natural logarithm (ln) or base-10 logarithm (log):

ln(1.5) = ln((1.019)^2t)

Using the logarithm property, we can move the exponent down in front:
ln(1.5) = 2t * ln(1.019)

Divide both sides by 2 * ln(1.019):
ln(1.5) / (2 * ln(1.019)) = t

Now, we can calculate t using a calculator:

t ≈ ln(1.5) / (2 * ln(1.019))
t ≈ 8.62 years

Therefore, it will take approximately 8.62 years for the investment to grow to $3000, assuming no withdrawals are made.