Let u = .

Then use the rule of logarithms  to write 

 =  = 

Let v = .

Then use the same rule of logarithms  to write 

 =  = 

So now the system is

u +  = 2
 + v = 

Clear each of fractions by multiplying the first thru by v and the
second one through by 3u

 uv + 1 = 2v
3 + 3uv = 8u

Solve the first for uv

 uv + 1 = 2v
     uv = 2v - 1

Substitute for uv in the second:

3 + 3uv = 8u
3 + 3(2v-1) = 8u

3 + 6v - 3 = 8u
        6v = 8u
        3v = 4u
         v = 
         
Substitute in

3 + 3uv = 8u

3 + 3u = 8u

Cancel 3's in the second term:

3 + u(4u) = 8u

3 + 4u² = 8u

    4u² - 8u + 3 = 0

(2u - 3)(2u - 1) = 0

2u - 3 = 0;      2u - 1 = 0
    2u = 3;          2u = 1
     u =             u = 
     
    3v = 4u          3v = 4u
    3v = 4        3v = 4 
    3v = 6           3v = 2
     v = 2            v =        

So we have (u,v) = (,2) or (u,v) = (,)

But we want x and y, not u and v, so we substitute back in each case:

(u,v) = (,2)
        
 u =          v = 
  =          2 =     

Write in exponential form using the rule:  is equivalent to 

x =  =  = 3³ = 27        y = 8² = 64

So one solution is (x,y) = (27,64)

For the other case:

(u,v) = (,)
        
 u =          v = 
  =           =     

Write in exponential form using the rule:  is equivalent to 

x =  =  = 3        y =  =  = 2² = 4

So the other solution is (x,y) = (3,4)

There are two solutions:

(x,y) = (27,64)  and   (x,y) = (3,4)




Edwin