Question

The area, Acm², of a patch of mould is measured daily. The area, n days after the measurements started, is given by the formula A = Asub0b^n". When n = 2, A = 1.8 and when n = 3, A = 2.4. Find the value of b.​

Answer 1

b = (4/3), with rounding

Explanation:

Area = A, in cm^2

n = days

A(n) =
`A_(0)`
`b^(n)`

------------------

Data:

n = 2, A = 1.8

A(n) =
`A_(0)`
`b^(n)`

1.8 =
`A_(0)`
`b^(2)`

n = 3, A = 2.4

2.4 =
`A_(0)`
`b^(3)`

Although we are not given the initial area, we can still make the calculation since we have two data points and each has the same initial area,
`A_(0)`. Lets take the first equation and multiply it by b, in order to bring the
`b^(2)` to
`b^(3)`

1.8 =
`A_(0)`
`b^(2)`

b(1.8 =
`A_(0)`
`b^(2)`)

1.8b =
`A_(0)`
`b^(3)`

We can know from the second equation that 2.4 =
`A_(0)`
`b^(3)` . We can eliminate this term in the above relationship by substituting 2.4:

1.8b =
`A_(0)`
`b^(3)`

1.8b = 2.4

b = 1.3333

or (4/3)

Check:

Does the value of (4/3) for b work for both data points? To check this, we do need to assume an initial starting area. Lets assume the starting area,
`A_(0)` = 1.

n = 2, A = 1.8, b = (4/3)

A(n) =
`A_(0)`
`b^(n)`

1.8 = 1*
`(4/3)^(n)`

1.8 = 1.78 Close enough?

n = 3, A = 2.4, b = (4/3)

2.4 =
`A_(0)`
`(4/3)^(n)`

2.4 = 1*
`(4/3)^(3)`

2.4 = 2.37 Close enough?