To determine the power function `y=ax^b` from the given coordinates: `(4,8) ` and `(8,30)` , we set-up system of equations by plug-in the values of `x` and `y` on `y=ax^b` .

Using the coordinate `(4,8)` , we let `x=4` and `y =8` .

First equation: `8 = a*4^b`

Using the coordinate `(8,30)` , we let `x=8` and `y =30` .

Second equation: `30 = a*8^b`

Isolate "a" from the first equation.

`8 = a*4^b`

`8/4^b= (a*4^b)/4^b`

`a= 8/4^b`

Plug-in` a=8/4^b` on `30 = a*8^b` , we get:

`30 = 8/4^b*8^b`

`30 = 8*8^b/4^b`

`30 = 8*(8/4)^b`

`30 = 8*(2)^b`

`30/8= (8*(2)^b)/8 `

`15/4=2^b`

Take the "ln" on both sides to bring down the exponent by applying the

natural logarithm property: `ln(x^n)=n*ln(x)` .

`ln(15/4) =ln(2^b)`

`ln(15/4) =b*ln(2)`

Divide both sides by `ln(2) ` to isolate b.

`(ln(15/4))/ln(2) =(b*ln(2))/(ln(2))`

`b =(ln(15/4))/ln(2) or 1.91` (approximated value).

Plug-in `b= 1.91` on `a=8/4^b` , we get:

`a=8/4^1.91`

`a~~ 0.566` (approximated value)

Plug-in `a~~0.566` and `b ~~ 1.91` on `y =ax^b` , we get the power function as:

`y =0.566x^1.91`

## See eNotes Ad-Free

Start your **48-hour free trial** to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Already a member? Log in here.