Rearranging economics equation - steady state of capital

Question

In my textbook, there are two formulas that follow on from each other, which I can't seem to be able to recreate. These are equations from the Solow-Swan model of growth in economics which state, that growth in capital goods is zero, when total new investment (new capital goods) is equal to depreciation in capital goods (capital no longer useful and eliminated from the stock)

So the first formula that highlights this is:

$s(K^*)^\alpha=\delta K^*$

But I am unsure of the steps to make $K^*$ the subject of this equation, so you can solve for capital in a general way, without any values for the variables.

$K^*= (\frac{\delta}{s})^\frac{1}{\alpha -1} = (\frac{s}{\delta})^\frac{1}{1-\alpha} $

I hope that I have given enough detail on the economic theory part of it, although, I cannot see how it is relevant to this problem.

Answer 1

Divide both sides of the first equation by $\delta$ to get $s(K^*)^\alpha/\delta=\delta K^*/\delta=K^*.$ Then divide both sides by $(K^*)^\alpha$ to get $s/\delta (K^*)^\alpha/(K^*)^\alpha=s/\delta=\delta (K^*)^{1-\alpha}$. Finally, take the $1-\alpha$th root of both sides to get the rightmost equality you've given. To get the first one, note that $(a/b)^c=(b/a)^{-c}$.

Answer 2

It seems that you have already got your answer in your question of; $K^*= (\frac{\delta}{s})^\frac{1}{\alpha -1} = (\frac{s}{\delta})^\frac{1}{1-\alpha} $

But anyway, the way to get reach that answer (if that is what you can't do), is by doing the following:

Divide both sides by $s$ and both sides by $k^*$, so you get:

$$\frac {(k^*)^\alpha}{k^*} = \frac {\delta}{s}$$

Due to index laws which state that $\frac {a^n}{a^m} = a^{n-m}$, so:

$$(k^*)^{\alpha - 1} = \frac {\delta}{s}$$ To get the $k^*$, by itself we use the fact $a^n = b$, then $a=b^{\frac {1}{n}}$, so:

$$\therefore k^* = (\frac {\delta}{s})^{\frac {1}{\alpha - 1}}$$ To get the most rightmost equation note that as @manofbear stated $(\frac {a}{b})^n = (\frac {b}{a})^{-n}$, so:

$$\therefore k^* = (\frac {\delta}{s})^{\frac {-1}{\alpha - 1}} = (\frac {\delta}{s})^{\frac {1}{1 - \alpha}}$$