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3 deleted 2 characters in body

As is said, this is too long for a comment, but no more substantial.

The related question in Jonas Meyer's comment seems to have settled almost anything involved, except that, as is common, the original person posting never supplied the claimed continued fraction.

I once read a book intended for high-school students that pointed out that $$a^b = b^a$$ is equivalent to $$\frac{\log a}{a} = \frac{\log b}{b}$$ and so any pair can be found by drawing a line through the origin that intersects the curve $y = \log x$ and taking $a,b$ as the $x$-coordinates of the two intersection points. This was given to motivate the idea that the only integral pair is $2^4 = 4^2.$ Of course there is also the point $(e,1)$ at which the line $y = x / e$ through the origin is tangent to the curve.

Are there any other pairs $(a,b)$ a,b$with both rational? 2 rational pairs As is said, this is too long for a comment, but no more substantial. The related question in Jonas Meyer's comment seems to have settled almost anything involved, except that, as is common, the original person posting never supplied the claimed continued fraction. I once read a book intended for high-school students that pointed out that $$a^b = b^a$$ is equivalent to $$\frac{\log a}{a} = \frac{\log b}{b}$$ and so any pair can be found by drawing a line through the origin that intersects the curve$ y = \log x$and taking$a,b$as the$x$-coordinates of the two intersection points. This was given to motivate the idea that the only integral pair is$2^4 = 4^2.$Of course there is also the point$(e,1)$at which the line$y = x / e$through the origin is tangent to the curve. Are there any other pairs$(a,b)$with both rational? 1 As is said, this is too long for a comment, but no more substantial. The related question in Jonas Meyer's comment seems to have settled almost anything involved, except that, as is common, the original person posting never supplied the claimed continued fraction. I once read a book intended for high-school students that pointed out that $$a^b = b^a$$ is equivalent to $$\frac{\log a}{a} = \frac{\log b}{b}$$ and so any pair can be found by drawing a line through the origin that intersects the curve$ y = \log x$and taking$a,b$as the$x$-coordinates of the two intersection points. This was given to motivate the idea that the only integral pair is$2^4 = 4^2.$Of course there is also the point$(e,1)$at which the line$y = x / e\$ through the origin is tangent to the curve.