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Find all distinct integers $x$ and $y$ that satisfy the following equation.

$$ x\log y=y\log x. $$

Obviously, if $x=y$, the equation is satisfied. I found $x=2$, and $y=4$. I think we cannot find all solutions (if there are many).

P.S. The base of the $\log(\cdot)$ is not important.

drzbir
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3 Answers3

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For positive real $x,y$ $$x\log y=y\log x\iff \frac1y\log(y)=\frac1x\log(x)\implies y^{\dfrac1y}=x^{\dfrac1x}$$

Now, show that $f(z)z=z^{\dfrac1z}$ is not constant

lab bhattacharjee
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Here using some of the $\log$ laws we can find all solution classes.

$$a \log (b)=\log(b^a)$$

and

$$e^{\log(x)}=x$$ shall be all that is required.

$$x\log y = y \log x$$

$$\log y^x = \log x^y$$ $$e^{\log (y^x)} = e^{\log (x^y)}$$ $$y^x = x^y$$

Once we have reached this point, we have a duplicate of this question.

Community
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We have $x^y = y^x$, with $0 < x < y$ (WLOG).

Let $y= rx$, where $r > 1$. Then $x^{rx} = (rx)^x$. Taking the natural logarithm of both sides, $rx \ln x = x (\ln r + \ln x)$. Dividing by $x$, and rearranging, $(r−1) \ln x = \ln r.$ (Notice that $r = 1$ would be a solution, from which $y = x$.) With $r > 1$, $\ln x = (\ln r) / (r−1).$ Taking exponentials,$ x = e^{(\ln r) / (r−1)} = (e^{\ln r})^{1/(r−1)}$. Therefore we have the parametric solution, $x = r^{1/(r−1)}, y = rx = r^{r/(r−1)}.$

If we set $r = 1 + 1/k$, where $k > 0$, we have $x = (1 + 1/k)^k$, $y = (1 + 1/k)^{k+1}$.

Since $r > 1, 1/(r−1) $is positive, and so $x > 1$, and therefore $y > 1$. (Notice that, if we allow $0 < r < 1, 1/(r−1) $and $r/(r−1)$ are negative, and so again $x$ and $y$ are greater than $1$. Of course, this must be the case, because $r = a$ and $r = 1/a$ yield essentially the same solution, with $x$ and $y$ swapped.)

We have shown that, for a rational solution, $k$ is an integer; that is $x = [(a + 1)/a]^a$. If $a > 1$, $a$ and $a + 1$ are relatively prime, and so $[(a + 1)/a]^a$ is a fraction in its lowest terms, with denominator $> 1$, and therefore not an integer. So the only integer solution is $x = 2, y = 4$.

Hope this helps. Correct me if I'm wrong!

You can refer here for more discussion on the same problem!

puru
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