Question

The amount of money invested in a certain account increases according to the following function, where is the initial amount of the investment, and is the amount present at time (in years).
y= yoe^0.035t

After how many years will the initial investment be doubled? Do not round any intermediate computations, and round your answer to the nearest tenth.

Answer 1

Answer:

Therefore the initial amount will be doubled after 19.8 years.

Step-by-step explanation:

Given that,

The amount of money increases according to the following function

$y=y_0e^{0.035t}$

where $y_0$ = the initial amount of the investment, and y = the amount present at time t (in years).

Let after t years the initial amount will be double.

$\therefore y=2 y_0$, t=t

$y=y_0e^{0.035t}$

$\Rightarrow 2y_0=y_0e^{0.035t}$

$\Rightarrow 2=e^{0.035t}$

Taking ln both sides

$\Rightarrow ln(2)=ln(e^{0.035t})$

$\Rightarrow ln(2)={0.035t}$                 [ $\because lne^a=a$]

$\Rightarrow t=\frac{ln(2)}{0.035}$

$\Rightarrow t=19.8$ years

Therefore the initial amount will be doubled after 19.8 years.