Algebra problem regarding system of logarithmic equations. The equations are:
$\log_{4}(x)+\log_{4}(y)=5$
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6$
I attempted to solve this problem by solving the pair of equations for $x$.
For the first equation:
$\Longrightarrow \log_{4}(xy)=5 \Longrightarrow xy=4^{5} \Longrightarrow xy=1024 \Rightarrow x=\dfrac{1024}{y}$
For the second equation:
$\Longrightarrow \log{4}(x)=\dfrac{6}{\log_{4}(y)} \Longrightarrow x=4^{\frac{6}{\log_{4}(y)}}$
Then,
$\Longrightarrow \dfrac{1024}{y}=4^{\frac{6}{\log_{4}(y)}}$
How should I move on from here?
 A: Let $u=\log_4(x)$, $v=\log_4(y)$. The equations can be written in $u$ and $v$
$$u+v=5,\ \ uv=6$$
The solutions are $u=2$, $v=3$, or $u=3$, $v=2$. Then you solve for $x$ and $y$
EDIT
As Bernard pointed out, $u$ and $v$ above are generally found as the solutions of the quadratic equation $t^2-5t+6=0$. But in this special case I just solved it by inspection.
Also, for completeness, $\log_4(x)=2$ implies $x=4^2=16$ and $\log_4(y)=3$ implies $y=4^3=64$. So the solutions are $x=16, y=64$ and $x=64, y=16$
A: We have:
$\log_{4}(x)+\log_{4}(y)=5$
$\Rightarrow \log_{4}(x)=5-\log_{4}(y) \hspace{5 mm}$ (i)
$\big(\log_{4}(x)\big)\big(\log_{4}(y)\big)=6 \hspace{4.25 mm}$ (ii)
Substituting (i) into (ii):
$\Rightarrow \big(5-\log_{4}(y)\big)\big(\log_{4}(y)\big)=6$
$\Rightarrow 5\log_{4}(y)-\big(\log_{4}(y)\big)^{2}=6$
$\Rightarrow \big(\log_{4}(y)\big)^{2}-5\log_{4}(y)+6=0$
$\Rightarrow \log_{4}(y)=\dfrac{-(-5)\pm\sqrt{(-5)^{2}-4(1)(6)}}{2(1)}$
$\Rightarrow \log_{4}(y)=\dfrac{5\pm{1}}{2}$
$\Rightarrow \log_{4}(y)=2,3$
$\Rightarrow y=16,64$
Using (i):
$\Rightarrow \log_{4}(x)=5-\log_{4}(16)$
$\hspace{19.5 mm}=5-2$
$\hspace{19.5 mm}=3$
$\Rightarrow x=64$
or
$\Rightarrow \log_{4}(x)=5-\log_{4}(64)$
$\hspace{19.5 mm}=5-3$
$\hspace{19.5 mm}=2$
$\Rightarrow x=16$
Therefore, the solutions to the system of equations is $x=16$ and $y=64$ or $x=64$ and $y=16$.
