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Cost Function Exponents Simplification

Hi everyone,

I'm working through an aspect of differential equation for an economics course and am stuck on a simplification aspect of it. The professor is assuming quite a bit of background mathematics knowledge, which is just slightly beyond my current level of understanding. If someone more skilled at these things could break down the following simplification on a variable by variable basis I'd be very grateful.


0=r−aB×w×A−1B×K−aB−1×Q1B0 = r-\frac{a}{B} \times w \times A^{-\frac{1}{B}} \times {K}^{-\frac{a}{B}-1} \times Q^{\frac{1}{B}}

Solving for K:

K−a+BB=A1B×Ba×rw×Q−1B{K}^{-\frac{a+B}{B}} = A^{\frac{1}{B}} \times \frac{B}{a} \times \frac{r}{w} \times Q^{-\frac{1}{B}}

Clearly, there's more to this problem than I've posted here but this is the part that I'm stuck on. I'm figuring this pretty much involves inverting the signs but if someone could give me a bit more of a detailed breakdown that would be great.

Thanks in advance for any replies.
Reply 1
Avatar for Ateo
Ateo
9
This is just the rules of indicies..

K−aB−1=raB⋅w⋅A−1B⋅Q1B\displaystyle K^{-\frac{a}{B} - 1} = \frac{r}{\frac{a}{B} \cdot w \cdot A^{-\frac{1}{B}} \cdot Q^{\frac{1}{B}}}

⇒K−a+BB=Ba⋅rw⋅A1B⋅Q−1B\Rightarrow \displaystyle K^{-\frac{a + B}{B}} = \frac{B}{a} \cdot \frac{r}{w} \cdot A^{\frac{1}{B}} \cdot Q^{-\frac{1}{B}}
Reply 2
Avatar for FishFullofRum
FishFullofRum
OP
Thanks for the response, but I was rather hoping for an explanation of the particulars of how those rules are applied in the above instance.
Reply 3
Avatar for davros
davros
16
Original post by FishFullofRum
Thanks for the response, but I was rather hoping for an explanation of the particulars of how those rules are applied in the above instance.


Well your first step should be to move all the terms except r to one side so that you have an equation that starts r = ....

Now see what factors you can multiply by that will help you to isolate the term with K in it.

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