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

Question

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Answer 1

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.