How long does it take to double a $1,000 investment that pays 6.5% annual interest, compounded monthly?

Compound interest formula: A=P (1+interst) ^n

P being the capital invested & n = the number of years, A the new capital

A = 1000 (1+0.065) ^n. They want to know n when the new capital is doubled,

2000=1000 (1.065) ^n & you will find n=11 years

2p=p (1+0.065/12) ^12t

2 = (1+0.065/12) ^12t

Solve for t

T=[log (2) / log (1+0.065/12) ]/12

T=10.7 years