`p (1 + R/100)^5` `= 2p ⇒ (1 + R/100)^5 = 2` Let `p (1 + R/100)^n` `= 8p.` Then, `(1 + R/100)^n = 8 = 2^3` `= {(1 + R/100)^5}^3` `= (1 + R/100)^15` `:. n = 15 years`.