Show using logarithms that the first equation can be transformed into the second.

Question

Show using logarithms that if $y^k = (1-k)zx^k(a)^{-1}$ then $y = (1-k)^{(1/k)}z^{(1/k)}x(a)^{(-1/k)}$.

Answer 1

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.

Answer 2

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?