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 $y = (1-k)^{(1/k)}z^{(1/k)}x(a)^{(-1/k)}$.
 A: The properties that you needs are:
$$
  \log ab=\log a \log b \qquad and \qquad\log a^b=b \log a
$$
Using these properties your expression become:
$$
\log y^k=k \log y= \log (1-k)zx^ka^{-1}=\log(1-k)+\log z+k\log x-1 \log a
$$
Now you can find $\log y$ as:
$$
\log y=\dfrac{1}{k}\left[\log(1-k)+\log z+k\log x-1 \log a\right]=
$$
$$
= \dfrac{1}{k}\log(1-k)+\dfrac{1}{k}\log z+ \log x-\dfrac{1}{k}\log a
$$
and, using the same properties, this becomes:
$$
\log y=\log \left[ (1-k)^{\frac {1}{k}}z^{\frac {1}{k}}xa^{-\frac {1}{k}}\right]
$$
now , exponentiating, you have the result.
But note that you can find the same result simply using the rules of radicals since from $y^k=A$ you have $y=\sqrt[k]{A}=A^{\frac{1}{k}}$.
In this way you see immediately that , if $K$ is even, we have a real solution only if your $A=(1-k)zx^ka^{-1}$ is positive. This condition is a bit more hidden using logarithms. You can see where it is request?
A: 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.
