# `(3,1) , (5,4)` Write an exponential function `y=ab^x` whose graph passes through the given points.

To determine the power function `y=ax^b` from the given coordinates: `(3,1)` and `(5,4)` , we set-up system of equations by plug-in the values of x and y on `y=ax^b` .

Using the coordinate `(3,1)` , we let `x=3` and `y =1` .

First equation: `1 = a*3^b`

Using the coordinate...

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To determine the power function `y=ax^b` from the given coordinates: `(3,1)` and `(5,4)` , we set-up system of equations by plug-in the values of x and y on `y=ax^b` .

Using the coordinate `(3,1)` , we let `x=3` and `y =1` .

First equation: `1 = a*3^b`

Using the coordinate `(5,4)` , we let `x=5` and `y =4` .

Second equation: `4 = a*5^b`

Isolate "`a` " from the first equation.

`1 = a*3^b`

`1/3^b= (a*3^b)/3^b`

`a= 1/3^b`

Plug-in `a=1/3^b` on `4 = a*5^b` , we get:

`4 = 1/3^b*5^b`

`4= 1*5^b/3^b`

`4= 1*(5/3)^b`

`4= (5/3)^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(4)=ln((5/3)^b)`

`ln(4)=b ln(5/3)`

Divide both sides by `ln(5/3)` to isolate `b.`

`(ln(4))/(ln(5/3))=(b ln(5/3))/(ln(5/3))`

`b =(ln(4))/(ln(5/3)) or 2.714` (approximated value)

Plug-in `b~~ 2.714` on `a=1/3^b` , we get:

`a=1/3^2.714`

`a~~0.051`

Plug-in `a~~0.051` and `b~~2.714` on `y=ax^b` , we get the power function as:

`y =0.051x^2.714`

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