compounded 2 times a year
ok, da equation yo want is
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
A=future amount
P=amount deposted or amount now
r=rate in decimal form
n=number of times a year compounded
t=time in years
so given
r=9%=0.09
n=2
t=t
and P and A are unknown
wait, we want ending to be triple begining
so A=3P
so
[tex]3P=P(1+\frac{0.09}{2})^{2t}[/tex]
[tex]3P=P(1+0.045)^{2t}[/tex]
[tex]3P=P(1.045)^{2t}[/tex]
divide both sides by P
[tex]3=(1.045)^{2t}[/tex]
take ln of both sides
[tex]ln(3)=ln((1.045)^{2t})[/tex]
[tex]ln(3)=2t(ln(1.045))[/tex]
divide both sides by 2ln(1.045)
[tex]\frac{ln(3)}{2ln(1.045)}=t[/tex]
use calculator
12.4794≈t
so about 12.5 years
