Finance word problem

Question

if $3000 is deposited into an account earning 8% compounded daily and at the same time $5000 is deposites into an account earning 5% compounded daily will be the first account ever be worth more than the second? If so when?

Answer 1

Hi Ana.

 

Compounding interest follows this formula:

 

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

 

A = amount in account

P = principal invested

r = rate of return

n = how many times a year it's compounded

t = time, typically years

 

If the first account overshoots the second account, there must be some time when they are equal to each other. Right after this happens is when the second account will be greater. Knowing this, you can equate the A for each account and solve for time.

 

A_8% = 3000*(1 + 0.08/365)^365*t

 

A_5% = 5000*(1 + 0.05/365)^365*t

 

You want A_8% to equal A_5%

 

So

 

3000*(1 + 0.08/365)^365*t = 5000*(1 + 0.05/365)^365*t

 

Solve for t by taking the natural log of each side. It's good to know these three identities before proceeding:

 

1. ln(a*b) = ln(a) + ln(b)

 

2. ln(a/b) = ln(a) - ln(b)

 

3. ln(a)^b = b*ln(a)

 

**

 

ln(3000*(1 + 0.08/365)^365*t) = ln(5000*(1 + 0.05/365)^365*t)

 

ln(3000) + ln(1 + 0.08/365)^365*t = ln(5000) + ln(1 + 0.05/365)^365*t

 

ln(3000) + 365*t*ln(1 + 0.08/365) = ln(5000) + 365*t*ln(1 + 0.05/365)

 

t*365*ln(1 + 0.08/365) - t*365*ln(1 + 0.05/365) = ln(5000) - ln(3000)

 

365*t*[ln(1 + 0.08/365) - ln(1 + 0.05/365)] = ln(5000/3000)

 

t = ln(5/3) / (365*[ln(1 + 0.08/365) - ln(1 + 0.05/365)])

 

t = 0.5108 / (365*(0.00021915 - 0.00013698))

 

t = 0.5108 / 0.02999205

 

t = 17 years

 

They'll each be worth just south of than $11,700 when the jump occurs. Seems like a long time to wait to me.

 

Hope this helps.