A question on indices I am solving a 7th grade text book and came across this question.
$81^x = 1/(125^y)$ where $x$ and $y$ are integers. Find $12xy$.
The answer given is $0$. While I can understand that when $x$ & $y$ are $0$ the above
equation is satisfied, I can't seem to figure how to arrive at the conclusion.
I tried something like:
--> (3^4x)(5^3y) = 1
--> 3^(4x + 3ky) = 1 where k is such that 3^k = 5
--> 4x + 3ky = 0
But with the above am not able to establish that $12xy$ is $0$.
Please help.
Thank you,
Ramana

If $(81)^x (125)^y = 1$, then it's clear that $x$ and $y$ can't have the same sign: If they're both positive, the left side is large, and if they're both negative, the right side is bigger. 
If $x$ is positive and $y$ is negative, then we can write
$$81^x = 125^{-y} \implies 3^{4x} = 5^{-3y}$$
But since $3$ and $5$ are prime numbers, it's necessarily true that $4x = 0 = -3y$. This is a consequence of unique factorization: No integer power of $5$ is equal to an integer power of $3$, except for $3^0 = 5^0$. The argument is easily adapted to the case $x < 0, y > 0$.
