Basic Logarithm question - I can't get both answers from quadratic Here's the Question : 
If  $xy$ = $64$ and $\log_x y + \log_y x = \frac{5}{2}$, find $x$ and $y$

I can get this to 
$$log_x y + \frac{1}{\log_x y} \frac{5}{2}$$
let $\log_x y = N$
$$N + \frac{1}{N} =  \frac{5}{2}$$
Multiply by 2
$$2N + \frac{2}{N} =  5$$
Multiply by N
$$2N^2 + 2 =  5N$$
$$2N^2 - 5N + 2 = 0$$
$$(2N - 1)(N - 2)$$
Giving : 
$$N = \frac{1}{2}$$
$$N = 2$$
Therefore : 
$$\log_x y = \frac{1}{2} $$
$$\log_x y = 2$$
Giving
$$x^2 = y$$
$$x^{\frac{1}{2}} = y$$
Part of the original question : 
$$xy = 64$$
As $x^2 = y$
$$x * x * x = 64$$
$$x^3 = 64$$
Therefore:
$$x = 4$$
$$y = 16$$

I can't seem to solve for $y = x^{\frac{1}{2}}$ though
Solving for $x^{\frac{1}{2}} = y$
$$x^{\frac{1}{2}} * x^{\frac{1}{2}} = 64$$
$$x^{\frac{1}{2} + \frac{1}{2}} = 64$$
$$x= 64$$
$$xy= 64$$
$$64y= 64$$
Therefore 
$$x = 64$$
$$y = 1$$
This is wrong though. 

Answer : 
$$(4,16) or (16,4)$$
I don't see how they got the second part. The first part makes sense but I'm not able to solve for $y = x^{\frac{1}{2}}$

$$\log_x(y)+\log_y(x)=\frac{\ln(y)}{\ln(x)}+\frac{\ln(x)}{\ln(y)}=\frac{\ln^2(x)+\ln^2(y)}{\ln(x)\ln(y)}=\frac{(\ln(x)+\ln(y))^2}{\ln(x)\ln(y)}-2=\frac{\ln^2(xy)}{\ln(x)\ln(y)}-2$$
Then if $xy=64$,
$$\log_x(y)+\log_y(x)=\frac{5}{2}\implies \frac{\ln^2(64)}{\ln(x)\ln(y)}=\frac{9}{2}\implies \frac{2\ln^2(64)}{9}=\ln(x)\ln(y)\underset{xy=64}{=}\ln(x)\ln(\frac{64}{x})=-\ln^2(x)+\ln(x)\ln(64)$$
and thus, if you set $X=\ln(x)$ you get 
$$X^2-\ln(64)X+\frac{2\ln^2(64)}{9}=0.$$
$$\Delta =\ln^2(64)-\frac{8}{9}\ln^2(64)=\frac{1}{9}\ln^2(64)$$
Therefore $$X=\frac{\ln(64)\pm\frac{1}{3}\ln(64)}{2}$$
and thus $X=\frac{1}{3}\ln(64)=\ln(\sqrt[3]{64})=\ln(4)$ ou $X=\frac{2}{3}\ln(64)=\ln(\sqrt[3]{64^2})=\ln(16).$
Therefore $x=4$ and $y=16$ or $x=16$ and $y=4$. Finally, we conclude that $(x,y)=(4,16)$ or $(x,y)=(16,4)$.
