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

  • $\begingroup$ Please use LaTeX. $\endgroup$ – Avi Steiner Jan 2 '14 at 3:52
  • $\begingroup$ As it is right now, I'm not sure I'm following your work. $\endgroup$ – Avi Steiner Jan 2 '14 at 3:58
  • $\begingroup$ @Ramana: After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Michael Albanese Jan 12 '14 at 8:21


$$(81)^x(125)^y=1\iff 3^{4x}\cdot 5^{3y}=3^0\cdot 5^0.$$

Notice that both $3$ and $5$ are prime numbers.

The number of each prime number in each side has to be the same.

P.S. A prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.


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 $x < 0, y > 0$.


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