Question 397408
a)  {{{2P = Pe^(0.015t)}}} ==> {{{2 = e^(0.015t)}}} ==> ln2  =0.015t 
==> t = ln2/0.015 = 46.2 years.

b)  {{{3P = Pe^(0.015t)}}} ==> {{{3 = e^(0.015t)}}} ==> ln3  =0.015t 
==> t = ln3/0.015 = 73.24 years.