log(z) = \(\displaystyle \displaystyle log\left (\left [y\right ]^{log(x)}\right )\)
\[{\log _b}(z) = {\log _b}(x){\log _b}(y)\]
From: \[{\log _b}{\rm{(}}{{\rm{x}}^a}) = a{\log _b}(x)\]
\[\begin{array}{l}
{\log _b}(z) = {\log _b}({y^{{{\log }_b}(x)}}) = > \\
z = {b^{({y^{{{\log }_b}(x)}})}}
\end{array}\]
Now a second formulation with a different result.
\[\begin{array}{l}
{\log _b}(z) = {\log _b}(x){\log _b}(y) = > \\
z = {b^{{{\log }_b}(x){{\log }_b}(y)}} = > \\
z = {b^{{{\log }_b}{{(x)}^{{{\log }_b}(y)}}}} = > \\
z = {x^{{{\log }_b}(y)}}
\end{array}\]
Are both results correct please? I believe I have verified they are equivalent but if so I have a further question about the laws of exponents that I would like to clear up, otherwise I would like to know what I am doing wrong. Also, are they in simplest form, can't get rid of the log function?
Finally, most textbook problems require simplifying a log expression to the log of a single argument before taking it's inverse ... I am now thinking that is not necessarily so ... true? false?
