Question 1104549
B is in the millions.  Just plugging into the formula,
{{{12=0.798(1.164)^t}}}

{{{log(12)=log(0.798)+log(1.164^t)}}}

{{{log(12)-log(0.798)=t*log(1.164)}}}

{{{t=(log(12)-log(0.798))/(log(1.164))}}}.  Compute this how you want.


t looks like about 17.8 years.