Question

How long will it take $5000 to grow to $6000 at an annual rate of 9% compounded monthly

Answer 1

To solve this problem, we can use the formula for compound interest:

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

where:
A = the amount of money in the account after t years
P = the principal amount (the initial investment)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years

In this case, we want to find the time it takes for an investment of $5000 to grow to $6000 at an annual rate of 9% compounded monthly. So we have:

P = $5000
A = $6000
r = 0.09 (9% annual interest rate)
n = 12 (compounded monthly)

We want to solve for t, the number of years. So we can rearrange the formula to solve for t:

t = (log(A/P)) / (n * log(1 + r/n))

Substituting the values we have:

t = (log(6000/5000)) / (12 * log(1 + 0.09/12))

Simplifying:

t = 3.10 years (rounded to two decimal places)

Therefore, it will take approximately 3.10 years (or about 37 months) for an investment of $5000 to grow to $6000 at an annual rate of 9% compounded monthly.

Answer 2

To solve this problem, we can use the formula for compound interest:

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

where:
A = the amount of money after t years
P = the principal amount (the initial investment)
r = the annual interest rate (as a decimal)
n = the number of times the interest is compounded per year
t = the number of years

We know that P = $5000, A = $6000, r = 0.09 (9%), and n = 12 (since interest is compounded monthly). We can solve for t algebraically:

A = P(1 + r/n)^(nt)
$6000 = $5000(1 + 0.09/12)^(12t)
$6000/$5000 = (1 + 0.09/12)^(12t)
1.2 = (1.0075)^(12t)
log(1.2) = log(1.0075)^(12t)
log(1.2) = 12t * log(1.0075)
t = log(1.2) / (12 * log(1.0075))
t ≈ 4.58

Therefore, it will take about 4.58 years (or about 55 months) for $5000 to grow to $6000 at an annual rate of 9% compounded monthly.