MATH 12 HELP NEEDED: Logarithms & Exponential Functions?

Question

Question: Calculate the number of years for an investment of $1000 to double at an interest rate of 7.2% for each compounding period.
a) annually b) semi-annually c) monthly d) daily

I know the equation for a) annually is: P = 1000(1.072)^Y which is 2x1000 = 1000(1.072)^Y
but I am having problem for coming up with the equations for b) semi-annually c) monthly d) daily

Thanks for all your helps =)

Answer 1

Compound interest formula is A= P(1+r/n)^nt where A is the new principle after t years. n is the number of times interest is calculated in one year, r is the rate of interest and P is the old principle.

Therefore, for your problems:

a) A = 2000, P = 1000, r = .072, n = 1 and you are solving for t.

b) A = 2000, P = 1000, r = .072, n = 2

c) n = 12

d) n = 365

Here is c) 2000 = 1000(1+.072/12)^(12t)
2 = (1.006)^(12t)
log 2 (base 1.006) = 12t
log 2 base 1.006/12 = t
log 2 (base 10)/log 1.006 (base 10)/12 = 9.6 yr

I'll let you do a), b) and d)

Answer 2

Use the same equation as the anual interest except to do semi-annual, divide the interest rate by 2 and double the number of years. For monthly divide the interest rate by 12 and 12 X the number of years, etc.

So for example, for 60 months, the equation would become.

1000(1+.072/12)^60

Answer 3

For semiannually n= 2
For monthly n= 12
For daily n = 365
For hourly n = 8760
For evey minute n = 525600

For monthly,
A = P ( 1 + r/n ) ^ nt , A= 2000 ( double up to 1000)
2000 = 1000 ( 1 + .072 / 12 )^12t
t = 9.6559 Years

For semiannually
A = P ( 1 + r/n ) ^ nt
2000 = 1000 ( 1 + .072 / 2 )^2t
t = 9.79931 yrs

For daily,
A = P ( 1 + r/n ) ^ nt
2000 = 1000 ( 1 + .072/365 )^365t
t = 9.62799 yrs.

Friend, that would help you. If not sorry for that.