Question

A quantity with an initial value of 830 grows exponentially at a rate such that the quantity doubles every 2 weeks. What is the value of the quantity after 21 day, to the nearest hundredth?

Answer 1

Answer:

The value of the quantity after 21 days is 2,347.59.

Step-by-step explanation:

The exponential growth function is

$A=A_0(1+r)^t$

A= The number of quantity after t days

$A_0$= initial number of quantity

r= rate of growth

t= time in days.

A quantity with an initial value of 830 grows at a rate such that the quantity doubles in 2 weeks = 14 days.

Now A= (2×830)= 1660

$A_0$ = 830

t = 14 days

r=?

Now plug all value in exponential growth function

$1660=830(1+r)^{14}$

$\Rightarrow \frac{1660}{830}= (1+r)^{14}$

$\Rightarrow 2= (1+r)^{14}$

$\Rightarrow (1+r) ^{14}=2$

$\Rightarrow (1+r)=\sqrt[14]{2}$

$\Rightarrow r=\sqrt[14]{2}-1$

Now, to find the quantity after 21 days, we plug $A_0$ = 830, t= 21 days in exponential function

$A=830( 1+\sqrt[14]{2}-1)^{21}$

$\Rightarrow A=830(\sqrt[14]2)^{21}$

$\Rightarrow A=830(2)^\frac{21}{14}$

$\Rightarrow A=830(2)^\frac{3}{2}$

$\Rightarrow A=2,347.59$

The value of the quantity after 21 days is 2,347.59.