Answer
The simplified solution set for the system of equations is $\left\{ \left( 8,2 \right) \right\}$.
Work Step by Step
We use the substitution method to solve the provided system of equations:
$\left\{ \begin{align}
& {{\log }_{y}}x=3....\text{(I)} \\
& {{\log }_{y}}\left( 4x \right)=5....\text{(II)}
\end{align} \right.$
By using the logarithm rule ${{\log }_{m}}\left( ab \right)={{\log }_{m}}a+{{\log }_{m}}b $ in equation (II), we get
$\begin{align}
& {{\log }_{y}}\left( 4x \right)=5 \\
& {{\log }_{y}}x+{{\log }_{y}}4=5 \\
& {{\log }_{y}}x=5-{{\log }_{y}}4
\end{align}$
Substitute this value of ${{\log }_{y}}x $ in equation (I),
$\begin{align}
& {{\log }_{y}}x=3 \\
& 5-{{\log }_{y}}4=3 \\
& {{\log }_{y}}4=2
\end{align}$
The above equation is in a single variable. To solve this equation, transform the equation to the power of y on both sides.
$\begin{align}
& {{\log }_{y}}4=2 \\
& {{y}^{{{\log }_{y}}4}}={{y}^{2}}
\end{align}$
We use the logarithm rule ${{m}^{{{\log }_{m}}a}}=a $ to simplify as shown below:
$\begin{align}
& {{y}^{{{\log }_{y}}4}}={{y}^{2}} \\
& 4={{y}^{2}} \\
& y=\pm 2
\end{align}$
The solution to this equation is $ y=2$ or $ y=-2$, but since the base of a logarithmic function cannot be a negative value, we ignore the solution $ y=-2$.
Putting the values of y in equation (I) to find values for x, we get,
For $ y=2$,
$\begin{align}
& {{\log }_{y}}x=3 \\
& {{\log }_{2}}x=3 \\
\end{align}$
To solve for $ x $ in this logarithmic equation, transform the equation to a power of 2 on both sides,
$\begin{align}
& {{\log }_{2}}x=3 \\
& {{2}^{{{\log }_{2}}x}}={{2}^{3}}
\end{align}$
Apply the logarithm rule ${{m}^{{{\log }_{m}}a}}=a $ to simplify,
$\begin{align}
& {{2}^{{{\log }_{2}}x}}={{2}^{3}} \\
& x={{2}^{3}} \\
& x=8
\end{align}$
Hence, for $ y=2$, we have $ x=8$.
