please help with these math problems?

interest compounded n times per annum:

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

interest compounded continuously:

A = P * e^(rt)

what more do you need ??

You're welcome !

Ω

The equation you need is:

A = Pe^(rt)

where P is the principal amount, r is the interest rate, t is the number of years, and A is the amount in "t" years.

Plug in what we know and solve for t:

3000 = 1000e^(0.07t)

3 = e^(0.07t)

Natural Log of both sides:

ln(3) = 0.07t

ln(3) / 0.07 = t

Using a calculator:

1.09861 / 0.07 = t

t = 15.6944

It will take 15.6944 years

---

For the part 1 of the other question, you need a different equation:

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

Part 1 and 2 asks to find out how long it takes to double at 8% interest when compounded quarterly and continuously. Part 1 first:

2400 = 1200(1 + 0.08/4)^(4t)

2 = (1 + 0.08/4)^(4t)

2 = (1 + 0.02)^(4t)

2 = (1.02)^(4t)

Log of base 1.02 on each side:

log-1.02(2) = 4t

t = log-1.02(2) / 4

t = log(2) / (4 log(1.02))

using a calculator and rounding to 3 decimal places:

8.751 years

---

Part 2, back to the continuously compounded equation from above:

A = Pe^(rt)

2400 = 1200e^(0.08t)

2 = e^(0.08t)

ln(2) = 0.08t

t = ln(2) / 0.08

t = 8.664

8.664 years (slightly less time than if compounded quarterly)