The client wants to know how long it will take for an initial deposit of $30000 to double at an annually compounded interest rate of 2.0%
What is the doubling time (assuming no withdrawals or additional deposits are made)?
1.02^n = 2
n = log2/log1.02
That is wrong
To determine the doubling time for an investment with compound interest, you can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A = final amount
P = principal (initial deposit)
r = annual interest rate (expressed as a decimal)
n = number of times interest is compounded per year
t = time in years
In this case, the principal (P) is $30,000, the annual interest rate (r) is 2.0% (or 0.02 as a decimal), and the investment is compounded annually (n = 1). We want to find the time (t) it takes for the principal to double, so the final amount (A) will be $60,000.
Let's rearrange the formula to solve for t:
A = P(1 + r/n)^(nt)
60,000 = 30,000(1 + 0.02/1)^(1*t)
Divide both sides of the equation by 30,000:
2 = (1 + 0.02)^t
Take the logarithm of both sides to solve for t:
log(2) = log(1 + 0.02)^t
Using the logarithmic property, we can bring down the exponent:
log(2) = t * log(1 + 0.02)
Now, divide both sides of the equation by log(1 + 0.02):
t = log(2) / log(1 + 0.02)
Calculating this using a calculator or computer software, we find:
t ≈ 35.0 years
Therefore, it will take approximately 35 years for the initial deposit of $30,000 to double at an annually compounded interest rate of 2.0%, assuming no withdrawals or additional deposits.