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

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

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 (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)^(21)/(14)`

`\Rightarrow A=830(2)^(3)/(2)`

`\Rightarrow A=2,347.59`

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