Question 566479
I'll connect you.
Let's change variables.
{{{w=log(9,x)}}} and {{{z=log(y,8)}}}
The first equation transforms easily:
{{{log(9,x)+log(y,8)=2}}} --> {{{w+z=2}}}
For the next one, we have to use a property of logarithms.
As with all the properties of logarithms, you can remember it, or you can rediscover it from the definition of logarithm every time you need it.
I always opted for the second choice, but in this case, rediscovery is not that easy, so I say understand the proof once, and then try to remember that
{{{log(a,b)=1/log(b,a)}}}
To help students remember that property, a popular tutor used to call it "skewer it and turn it upside down." That helped because he would write a very large {{{1}}} above the {{{log(b,a)}}} and pretend it was a toothpick skewering an olive, before pretending to turn it upside down and take it to his mouth, and 
the {{{log(a,b)}}} he would then write on the board sort of looks like an upside down {{{log(b,a)}}}. So
{{{log(x,9)=1/log(9,x)}}} , so {{{log(x,9)=1/w}}} and
{{{log(8,y)=1/log(y,8)}}} , so {{{log(8,y)=1/z}}}
With that, the second equation gets transformed
{{{log(x,9)+ log(8,y)=8/3}}} --> {{{1/w+1/z=8/3}}} --> {{{(w+z)/wz=8/3}}}
So we have two equations
{{{w+z=2}}} and {{{(w+z)/wz=8/3}}}
and just need to find w and z. No logarithm worries (for now).
At this point I would divide the first equation by the second equation
{{{(w+z)/(((w+z)/wz))}}}={{{2/((8/3))}}} --> {{{(w+z)*(wz/(w+z))=2*(3/8)}}} --> {{{wz=3/4)}}}
If you figure out here that the numbers {{{w}}} and {{{z}}} must be {{{3/2}}} and {{{1/2}}}, good for you.
I had to remember that they would be solutions of the quadratic equation
{{{x^2-2x+3/4=0}}} <--> {{{4x^2-8x+3=0}}} and use the quadratic formula.
It could be that {{{w=3/2}}} and {{{z=1/2}}} ,
or that {{{w=1/2}}} and {{{z=3/2}}} .
The first solution:
{{{w=3/2}}} and {{{z=1/2}}}
{{{w=3/2}}} --> {{{log(9,x)=3/2}}} --> {{{x=9^(3/2)=(sqrt(9))^3=3^3=27}}}
and {{{z=1/2}}} --> {{{log(8,y)=1/z=1/(1/2)=2}}} --> {{{y=8^2=64}}}
The second solution:
{{{w=1/2}}} and {{{z=3/2}}}
{{{w=1/2}}} --> {{{w=log(9,x)=1/2}}}--> {{{x=9^(1/2)=sqrt(9)=3}}}
{{{z=3/2}}} --> {{{log(8,y)=1/z=1/(3/2)=2/3}}} --> {{{y=8^(2/3)=(root(3,8))^2=2^2=4}}}