How do I solve for X and Y when Y = log_3(6) and X = log_6(5)? The question asks me to express log_3(10) in terms of Y and X. What do they mean when they say "express"?

Hello!
"Express" something in terms of X and Y means to find an expression (formula, rule) that gives this "something" as a result of operations on X and Y. The example of a formula is  X+2Y.
Denote the number in question  log_3(10)  as Z.
Then we need to "solve for Z", not "solve for X and Y".
To do this, we need some properties of logarithms:
log_a(b*c) = log_a(b) + log_a(c),  (logarithm of a product)
log_a(b/c) = log_a(b) - log_a(c),  (logarithm of a quotient)
log_a(a) = 1,
log_b(a) = (log_c(a))/(log_c(b))  (change of a base),
log_a(b) = 1/(log_b(a))   (a consequence of change of a base).
Then we can state that:
log_3(10) = log_3(2*5) =   (log of a product)
= log_3(2) + log_3(5) = log_3(6/3) + log_3(5)   (log of a quotient). Also I am rewriting 2 as 6/3 since 2 = 6/3
=log_3(6) - log_3(3) + log_3(5) =   (change of a base)
= Y - 1 + (log_6(5))/(log_6(3)) = Y - 1 + X*log_3(6) =   
= Y - 1 + X*Y.
This is the expression we need.
http://dl.uncw.edu/digilib/Mathematics/Algebra/mat111hb/EandL/logprop/logprop.html