Show using logarithms that the first equation can be transformed into the second. Show using logarithms that if $y^k = (1-k)zx^k(a)^{-1}$ then $y = (1-k)^{(1/k)}z^{(1/k)}x(a)^{(-1/k)}$.

Take logarithm of both sides of the equation $\,y^k =  \left(1-k\right)\,z\,x^k\,\left(a\right)^{-1},\,$ get
$$
\begin{aligned}
y^k = \left(1-k\right)\,z\,x^k\,a^{-1}  
&\implies
\ln\left( y^k \right) = \ln\left(\left(1-k\right)\,z\,x^k\,\left(a\right)^{-1}  \right) 
\\
&\implies
k\ln y = \ln\left(1-k \right) + \ln z  + \ln\left(x^k \right) -\ln a
\\
&\implies
k\ln y = \ln\left(1-k \right) + \ln z  + k\ln x -\ln a
\\
&\implies
\ln y = \frac{1}{k}\ln\left(1-k \right) + \frac{1}{k}\ln z  + \frac{k}{k}\ln x-\frac{1}{k}\ln a
\\
&\implies
\ln y = \ln\left(\left(1-k \right)^\frac{1}{k}\right) + \ln \left(z^\frac{1}{k}\right)  + \ln x -\ln \left(a^\frac{1}{k}\right)
\\
&\implies
\ln y = \ln\left(\left(1-k \right)^\frac{1}{k} \,z^\frac{1}{k}\, x \,a^\frac{-1}{k}\right)
\\
&\implies
y = \left(1-k \right)^\frac{1}{k} \,z^\frac{1}{k}\, x \,\left(a\right)^\frac{-1}{k}
\end{aligned}
$$
Q.E.D.
