Solving x^y=y^x with change of base of logarithm

Question

This guy solves $x^y=y^x$ by introducing a parameter $t$:

$$x=t^\frac{1}{t-1}$$

$$y=t^\frac{t}{t-1}$$

Fine, but my first instinct was to use the change-of-base formula:

$$\frac{\ln{y}}{\ln{x}} = \frac{\ln{x}}{\ln{y}}$$

$$(\ln{y})^2 = (\ln{x})^2$$

$$\ln{y} = \pm\ln{x}$$

given $y\ne x$,

$$\ln{y} = -\ln{x}$$

Why is this wrong?

How specifically did you go from $x^y =y^x$ to $\frac{\ln y}{\ln x}=\frac{\ln x}{\ln y}$? — Mike Earnest, Sep 17 '18 at 21:19
In addition to your your instinct, you can find almost all ideas for solving it here, because this equation has been discussed a lot already, e.g. here, here, etc. — Dietrich Burde, Sep 17 '18 at 21:22

score 1 · Accepted Answer · answered Sep 17 '18 at 21:20

1

$$x^y = y^x$$

does not imply

$${\ln(y)\over \ln(x)} = {\ln(x)\over \ln(y)} $$

but instead

$$y\ln(x) = x\ln(y)$$ or $${\ln(x)\over \ln(y)} = {x\over y}$$

answered Sep 17 '18 at 21:20

David P

1

Note that the logarithm is only defined for positive numbers. – katosh Sep 17 '18 at 21:28
@katosh not true at all. Branch cuts: use em. In particular, use the various branches of the Lambert W to solve on $\mathbb C$ – Brevan Ellefsen Nov 21 '18 at 04:01

1 Answers1