<s>[INST] 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 $y = x^{\frac{1}{2}}$
 [/INST] If $$y=x^{1/2}$$ and $$xy=64$$ then $$x^{3/2}=64 = 2^6$$ so  $$x=2^{6\times 2/3} = 2^4 =16$$ and $$y=16^{1/2}=4.$$  
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