We are asked to write the equation for a power function whose graph passes through the points (4,3) and (8,15).
We substitute the known values of x and y into the basic equation to get two equations with two unknowns (a and b) and then solve the system for the coefficients.
`3=a*4^b, 15=a*8^b`
Solving the first equation for a we get:
`a=3/(4^b)`
Substitute this expression for a in the second equation to get:
`15=3/(4^b)*8^b`
`15=3*(8/4)^b`
...
We are asked to write the equation for a power function whose graph passes through the points (4,3) and (8,15).
We substitute the known values of x and y into the basic equation to get two equations with two unknowns (a and b) and then solve the system for the coefficients.
`3=a*4^b, 15=a*8^b`
Solving the first equation for a we get:
`a=3/(4^b)`
Substitute this expression for a in the second equation to get:
`15=3/(4^b)*8^b`
`15=3*(8/4)^b`
`2^b=5`
So `b=(ln(5))/(ln(2))~~2.322`
Now substitute for b to get a:
`a=3/(4^b)=3/(4^((ln(5))/(ln(2))))=3/25=.12`
So the model is `y=3/25x^((ln(5))/(ln(2)))"or" y~~.12x^2.322`
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