Question

Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required for the amount to (a) double and (b) triple.
r = 0.085

Answer 1

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

                $A = Pe^{rt}$

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

               A = 2 × $3000

               A = $6000

               $6000 = 3000e^{0.085t}$

               $\frac{6000}{3000} = e^{0.085t}$

               $2 = e^{0.085t}$

Take $log_{e}$ of both sides

               $log_{e}2 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

            ∴ $ln2 = 0.085t$

               $t = \frac{ln2}{0.085}$

               $t = \frac{0.693}{0.085}$

               t = 8.15

(b) For the amount to double, A = 3 × P

               A = 3 × $3000

               A = $9000

               $9000 = 3000e^{0.085t}$

               $\frac{9000}{3000} = e^{0.085t}$

               $3 = e^{0.085t}$

Take $log_{e}$ of both sides

               $log_{e}3 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

            ∴ $ln3 = 0.085t$

               $t = \frac{ln3}{0.085}$

               $t = \frac{1.0986}{0.085}$

               t = 12.92