Question 549740
{{{log(9,x)}}} + {{{log(y,8)}}} = 2
{{{log(x,9)}}} + {{{log(8,y)}}} = {{{8/3}}}
<pre> 
Let u = {{{log(9,x)}}}.

Then use the rule of logarithms {{{log(B,A)=1/log(A,B)}}} to write 

{{{log(x,9)}}} = {{{1/log(9,x)}}} = {{{1/u}}}

Let v = {{{log(8,y)}}}.

Then use the same rule of logarithms {{{log(B,A)=1/log(A,B)}}} to write 

{{{log(y,8)}}} = {{{1/log(8,y)}}} = {{{1/v}}}

So now the system is

u + {{{1/v}}} = 2
{{{1/u}}} + v = {{{8/3}}}

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 = {{{(4u)/3}}}
         
Substitute in

3 + 3uv = 8u

3 + 3u{{{((4u)/3)}}} = 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 = {{{3/2}}}            u = {{{1/2}}}
     
    3v = 4u          3v = 4u
    3v = 4{{{(3/2)}}}        3v = 4{{{(1/2)}}} 
    3v = 6           3v = 2
     v = 2            v = {{{2/3}}}       

So we have (u,v) = ({{{3/2}}},2) or (u,v) = ({{{1/2}}},{{{2/3}}})

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

(u,v) = ({{{3/2}}},2)
        
 u = {{{log(9,x)}}}         v = {{{log(8,y)}}}
 {{{3/2}}} = {{{log(9,x)}}}         2 = {{{log(8,y)}}}    

Write in exponential form using the rule: {{{A=log(B,C)}}} is equivalent to {{{C=B^A}}}

x = {{{matrix(2,1,"",9^(3/2))}}} = {{{(sqrt(9))^3}}} = 3³ = 27        y = 8² = 64

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

For the other case:

(u,v) = ({{{1/2}}},{{{2/3}}})
        
 u = {{{log(9,x)}}}         v = {{{log(8,y)}}}
 {{{1/2}}} = {{{log(9,x)}}}         {{{2/3}}} = {{{log(8,y)}}}    

Write in exponential form using the rule: {{{A=log(B,C)}}} is equivalent to {{{C=B^A}}}

x = {{{matrix(2,1,"",9^(1/2))}}} = {{{sqrt(9)}}} = 3        y = {{{matrix(2,1,"",8^(2/3))}}} = {{{(root(3,8))^2}}} = 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</pre>