Question 151295
{{{A=P(1+r/n)^(nt)}}} Start with the compound interest formula.



{{{1500=1000(1+0.12/12)^(12t)}}} Plug in {{{A=1500}}}, {{{P=1000}}}, {{{r=0.12}}} (note: 12%=0.12) and {{{n=12}}}



{{{1500/1000=(1+0.12/12)^(12t)}}} Divide both sides by 1000.



{{{3/2=(1+0.12/12)^(12t)}}} Reduce



{{{3/2=(1+0.01)^(12t)}}} Divide {{{0.12/12}}} to get 0.01



{{{3/2=(1.01)^(12t)}}} Add 1 to 0.01 to get 1.01



{{{log(10,(3/2))=log(10,((1.01)^(12t)))}}} Take the log of both sides.



{{{log(10,(3/2))=12t*log(10,(1.01))}}} Rewrite the right side using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{log(10,(3/2))/(12log(10,(1.01)))=t}}} Divide both sides by {{{12log(10,(1.01))}}}.



So the answer is {{{t=log(10,(3/2))/(12log(10,(1.01)))}}} which approximates to {{{t=3.39574}}}



So it takes about 3.4 years (or 40.8 months) to go from $1,000 to $1,500