Variable with an exponent variable

Question

I'm actually dealing with an economics problem, but it seems like the math is always what messes me up. Ignoring what the variables mean, I'm trying to understand how to get from step 1 to step 2. $$ \begin{align} (d+g)k_e^* &= sk_e^{*\alpha}\\ k_e^* &= \left(\frac{s}{d+g}\right)^\frac{1}{1-\alpha} \end{align} $$

Any input is appreciated

Answer 1

First, divide both sides by $(d+g)$: $$(d+g)k_e^* = sk_e^{*\alpha} \implies k_e^* = \frac{s}{(d+g)}k_e^{*\alpha}$$ Now, divide both sides by $k_e^*$, and use exponent rules to simplify: $$k_e^* = \frac{s}{(d+g)}k_e^{*\alpha} \implies \frac{k_e^*}{k_e^{*\alpha}} = \frac{s}{(d+g)} \implies {k_e^*}^{1-\alpha}= \frac{s}{d+g}$$ Now, raise both sides to the $1/(1-\alpha)$ power: $${k_e^*}^{1-\alpha}= \frac{s}{d+g}\implies \left({k_e^*}^{1-\alpha}\right)^{1/(1-\alpha)}= \left(\frac{s}{d+g}\right)^{1/(1-\alpha)}$$ Using exponent rules to simplify the right hand side, we have: $$k_e^* = \left(\frac{s}{d+g}\right)^\frac{1}{1-\alpha}$$

Answer 2

I give you a step-by-step derivation $$(d+g)k=sk^\alpha\\ \frac{k}{k^\alpha}=\frac{s}{d+g} \\ kk^{-\alpha}=\frac{s}{d+g} \\ k^{1-\alpha}=\frac{s}{d+g} \\ \left(k^{1-\alpha}\right)^{\frac{1}{1-\alpha}}=\left(\frac{s}{d+g}\right)^{\frac{1}{1-\alpha}} \\ k^{\frac{1-\alpha}{1-\alpha}}=\left(\frac{s}{d+g}\right)^{\frac{1}{1-\alpha}} \\ k=\left(\frac{s}{d+g}\right)^{\frac{1}{1-\alpha}}.$$