Homogeneous Function

[know]

examine the homogeneity of the function:

ln f (a,b,c)=-3,2*ln a +0,2 ln b +0,3 c +1

 

[Dr.Peterson]

examine the homogeneity of the function:

ln f (a,b,c)=-3,2*ln a +0,2 ln b +0,3 c +1

Please show us what you've tried, and where you are stuck:

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[know]

The first problem is that there is: ln y=...
Formula: f(λx1, λx2, . . . , λxn) = λ ^α f(x1, x2, . . . , xn).
How to get result? What I have to do first?

 

[topsquark]

examine the homogeneity of the function:

ln f (a,b,c)=-3,2*ln a +0,2 ln b +0,3 c +1

If you are "lost" then check a value.

If f(a, b, c) is homogeneous then we know that [math]f( \lambda a, \lambda b, \lambda c ) = \lambda ^{ \alpha }f(a, b, c)[/math] for some [math]\alpha[/math].

So let [math]\lambda = 2[/math]. Does
[math]f(2a, 2b, 2c) = -3.2 ~ ln(2a) + 0.2 ~ ln(2b) + 0.3 (2c) + 1 = 2^{ \alpha } (-3.2 ~ ln(a) + 0.2 ~ ln(b) + 0.3 c + 1)[/math]for some value of [math]\alpha[/math]?

-Dan

 

[Dr.Peterson]

The first problem is that there is: ln y=...
Formula: f(λx1, λx2, . . . , λxn) = λ ^α f(x1, x2, . . . , xn).
How to get result? What I have to do first?

You could start in several different ways.

One would be to raise e to the quantities on both sides of your given equation, so that it directly defines f: [MATH]e^{ln f (a,b,c)}=e^{-3,2 \ln a +0,2 \ln b +0,3 c +1}[/MATH]. Simplify that, and then apply the definition of homogeneous function to it.

Another would be to take the natural log of each side of your formula for a homogeneous function, to see what your function needs to do in the form it is presented.