Answer
9.9 years
Work Step by Step
$A=2000$
$P=1000$
$r=.07$
$n=12$
$A=P(1+(r/n))^{n*t}$
$2000 = 1000*(1+(.07/12))^{12*t}$
$2000/1000 = 1000*(1+(.07/12))^{12*t}/1000$
$2= (1+(.07/12))^{12*t}$
$2= (1+(.005833))^{12*t}$
$2= (1+.005833)^{12*t}$
$2=1.005833^{12*t}$
$ln 2 = ln 1.005833^{12*t}$
$ln 2 = 12*t*ln 1.005833$
$.693147 = t*12*.005816$
$.693147 = t*.06979$
$.693147/.06979 = t*.06979/.06979$
$9.93 =t$
9.9 years
If the interest rate and frequency of compounding are the same, then the time required to double the money doesn’t change.
