if $\log_xa+\log_ya=4\log_{xy}a$ prove that $x=y$ Let $x,y$ be numbers in the interval $(0,1)$ with the property that there exists a positive number $a\ne1$ such that 
$$\log_xa+\log_ya=4\log_{xy}a$$
I have used the property
$$\log _ba=\frac{\log a}{\log b}$$
and I have
$$\log_xa+\log_ya=\frac{\log a}{\log x}+\frac{\log a}{\log y}$$
$$\log a\left(\frac{\log x+\log y}{\log x \log y}\right)=4\frac{\log a}{\log(xy)} \implies (\log(xy))^2=4 \log x \log y$$
I don't see how this implies $x=y$
 A: Continuing your work...
$$(\log{x y})^2 = (\log{x}+\log{y})^2 = (\log{x})^2 + 2 \log{x} \, \log{y} + (\log{y})^2 = 4 \log{x} \, \log{y}$$
which means that
$$(\log{x})^2 - 2 \log{x} \, \log{y} + (\log{y})^2 = (\log{x} - \log{y})^2 = 0$$
which means that
$$\log{x} - \log{y} = 0 \implies x=y$$
A: $$\log_xa+\log_ya=4\log_{xy}a$$
$$\dfrac{1}{\log_ax}+\dfrac{1}{\log_ay}=4\dfrac{1}{\log_a{xy}}$$
$$\dfrac{{\log_ay}+{\log_ax}}{{\log_ax}\cdot{\log_ay}}=4\dfrac{1}{{\log_ay}+{\log_ax}}$$
$$\dfrac{({\log_ay}+{\log_ax})^2}{{\log_ax}\cdot{\log_ay}}=4$$
$$({\log_ay})^2+({\log_ax})^2+2\cdot\log_ax\cdot\log_ay=4{{\log_ax}\cdot{\log_ay}}$$
$$({\log_ay})^2+({\log_ax})^2-2\cdot\log_ax\cdot\log_ay=0$$
$$({\log_ay}-{\log_ax})^2=0$$
$$({\log_ay}-{\log_ax})=0$$
$${\log_ay}={\log_ax}$$
$$x=y$$
A: HINT: From $(\log xy)^2=4 \log x \log y$, write out the right hand side as
$$(\log x + \log y)^2=  (\log x)^2 + 2 \log x \log y + (\log y)^2 \; .$$
A: Continuing from where you left off, let $p=\log x$ and $q=\log y$. Then:
$$ \begin{align*}
(\log xy)^2 &= 4 \log x \log y \\
(\log x + \log y)^2&= 4 \log x \log y \\
(p+q)^2&= 4 pq \\
p^2 + 2pq + q^2 &= 4 pq \\
p^2 - 2pq + q^2 &= 0 \\
(p-q)^2 &= 0 \\
p-q &= 0 \\
p &= q \\
\log x &= \log y\\
x &= y
\end{align*} $$
as desired.
