Posted by **catherine** on Sunday, September 2, 2012 at 2:34pm.

how to count how long of $5000 at 7% interest compounded for the investment to increase to 10000?

it will be awesome if you can put the way

- math -
**Henry**, Tuesday, September 4, 2012 at 8:52pm
Use the maximum compounding frequency which is daily. This will minimize the time required to reach $10,000.

P = Po(1+r)^n.

P = $10,000.

Po = $5,000 = Initial deposit.

r = (7%/360) / 100%=0.0001944.= Monthly % rate expressed as a decimal.

n = The # of compounding periods.

5000(1.0001944)^n = 10000

Divide both sides by 5000:

(1.0001944)^n = 2

Take log of both sides:

n*Log(1.0001944) = Log2

n = Log2 / Log(1.0001944) = 3565 Compounding periods.

T = 3565comp. * iyr/360Comp = 9.9 yrs.

## Answer this Question

## Related Questions

- math - find the accumulated value of an investment of $10000 for 5 years at an ...
- precalc - A $5000 investment earns 7.2% annual interest, and an $8000 investment...
- Functions (Math) - In the formula A=p(1+i)^n, I can not remember how to sub a ...
- math - what formulas do i use for this: Investments Suppose $10,000 is invested ...
- pre calculus - A sum of $5000 is be invested in a bank. if the annual interest ...
- algebra- Help fast - use the compound interest formula A=P(1+r/n)^nt and A=Pe^rt...
- Math - For a savings account paying 2.4% in which interest is compounded MONTHLY...
- math - An amount of $5000 is invested at an interest rate of 7% per year, ...
- math - How long will it take for an investment of $13,000 to double if the ...
- math/ compounded - Scenario: A client comes to you for investment advice on his...