Find integer solutions for m - n(log3(2)) = 10(log9(6)).

From the properties of logarithms (log_ba = log_ca / log_cb), 10log₉6 can be rewritten as 10(log₃6 / log₃9). Since log₃9 = 2, 10log₉6 = 5log₃6. We then bring both log terms to the same side of the equation: m = 5log₃6 + nlog₃2. Again, from log properties (a(log_cb) = log_cb^a), this can be rewritten as m = log₃6⁵ + log₃2ⁿ. Since the logarithms have the same base, we can combine them using another log property (log_ab + log_ac = log_abc). This yields m = log₃6⁵2ⁿ. We can factor out the 2⁵ from the 6⁵ to obtain m = log₃3⁵2⁵2ⁿ. Combining the 2s: m = log₃3⁵2^{5 + n}. We then raise 3 to the power of both sides to get an equation without logarithms: 3^m = 3⁵2^{5 + n}. We can write an invisible 2⁰ term on the left side without changing the equation, giving 3^m2⁰ = 3⁵2^{5 + n}. Since m and n are integers, m = 5 and 5 + n = 0, meaning n = -5. So the solution is m = 5, n = -5.