In a savings account, a principal of $1000 is deposited at 5% per annum. In the second account a principal of $500 is deposited at 10% per annum. Both accounts are compounded continuously.

a. estimate the doubling time for each

we don't really need the amount

at first rate:
2 = 1 e^(.05t)
ln2 = .05t
t = 13.86 ---> the doubling time

at 2nd rate:
2 = 1 e^(.1t)

you finish it.

To estimate the doubling time for each account, we can use the formula for continuous compound interest:

A = P * e^(rt)

Where:
A = final amount
P = principal amount (initial deposit)
e = Euler's number (approximately 2.71828)
r = annual interest rate
t = time in years

Let's calculate the doubling time for each account:

1. Account #1:
P1 = $1000
r1 = 5% = 0.05

To estimate the doubling time, we need to find the time (t1) when the final amount (A1) is double the principal amount.

2P1 = P1 * e^(r1 * t1)
2 = e^(0.05 * t1)

To solve for t1, we can take the natural logarithm (ln) on both sides:

ln(2) = ln(e^(0.05 * t1))
ln(2) = 0.05 * t1

Divide both sides by 0.05:

t1 = ln(2) / 0.05
t1 ≈ 13.86 years

Therefore, it will take approximately 13.86 years for Account #1 to double.

2. Account #2:
P2 = $500
r2 = 10% = 0.10

Following the same steps as before:

2P2 = P2 * e^(r2 * t2)
2 = e^(0.10 * t2)

ln(2) = ln(e^(0.10 * t2))
ln(2) = 0.10 * t2

Divide both sides by 0.10:

t2 = ln(2) / 0.10
t2 ≈ 6.93 years

Therefore, it will take approximately 6.93 years for Account #2 to double.