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Given $2\log x^3y=6+3\log y-\log x$, x and y are positive integers. Show that $xy=100$. I have tried until $x^7=10^6 y$. Now, my problem is how to prove $x=y$.

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    $\begingroup$ There must be some mistake - you can go no further than $x^7=10^6y$. The equation will hold for any $(x,y)$ that satisfy this condition. $\endgroup$
    – Galc127
    May 17, 2016 at 15:26

2 Answers 2

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The claim is not generally true. As a counterexample, the given equation is solved by $x=10^2$, $y=10^8$:

$2\log\left((10^2)^3\cdot 10^8\right) = 2\log\left(10^6\cdot 10^8\right) = 2\cdot 14 = 28 = 6 + 24 - 2 = 6 + 3\log 10^8 - \log 10^2$

Indeed, this equation should be solved by any $(x,y)$ satisfying your equation $x^7 = 10^6 y$, as long as x and y are positive. This puts $xy = \dfrac{x^8}{10^6}$, which only equals $100$ in the case where $x=10$.

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Given that $x$ and $y$ are supposed to be integers, this is a number theory problem. My guess is that you should write the decomposition of $x$ and $y$ as products of prime numbers, i.e. $x = p_1^{\alpha_1}... p_r^{\alpha_r}$ and $y = q_1^{\beta_1}... q_s^{\beta_s}$, with $r, s \geq 1$, $p_1, p_2, ... p_r$ distinct prime numbers, same for $q_1, ... , q_s$, and exponents $\alpha_1, ... \alpha_r, \beta_1, ... \beta_s \geq 1$.

Once you have done that, see what the equation $x^7 = 10^6 y$ implies.

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