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.