Answer
$y=2, x=8$.
Work Step by Step
We are given the system:
$\begin{cases}
\log_{y}x=3\\
\log_{y}4x=5
\end{cases}$
Using logarithmic Inverse property of Exponent rules, $b^{\log_{b}a}=a$.
we can simplify the equation as follows by raising the equation to $y$.
$\begin{cases}
y^{\log_{y}x}=y^3\\
y^{\log_{y}4x}=y^5
\end{cases}$
$-- -- -- $
$\begin{cases}
x=y^3\\
4x=y^5
\end{cases}$
We use the Addition method. Mlutply Equation 1 by -4 and add it to Equation 2 to eliminate $x$.
$\begin{cases}
-4x+4y^3=0\\
4x-y^5=0
\end{cases}$
$-- -- -$
$4y^3-y^5=0,\\
y^3(4-y^2)=0,
y^3=0\\ y=0\\ or 4-y^2=0\\y=\pm 2$.
since $y$ is the base for logarithm. log of negative and 0 bases are undefined. only $y=2$ is relevant.
Therefore, substituting $y$ as a base for log and using logarithm's Inverse property of exponent rules, raising both sides of the equation to $2$, we can solve for $x$.
$\begin{cases}
2^{\log_{2}x}=2^3\\
2^{\log_{2}4x}=2^5
\end{cases}$
$-- -- --$
$x=8\\$
$4x=32$
$x=8$.
Therefore, the solution is $y=2, x=8$.
