I need to invert this function:

$$ y=\frac{\ln(x)}{\ln(x-1)}+1 $$

The domain is real (for x>1 and x!=2)

Why can't we just divide it like this: $$ y=ln(x-(x-1))+1 $$ and then it's: $$ y=ln(1)+1 $$ so it seems wrong. Where did I make the mistake?

  • $\begingroup$ Just out of curiosity: where did you come across this problem? $\endgroup$ – Thomas Nov 12 '12 at 16:42

In general, $\dfrac{\ln a}{\ln b}\ne \ln(a-b)$.

Remarks: $1.$ The false simplification was probably motivated by $\ln\left(\frac{a}{b}\right)=\ln a-\ln b$, which is true for positive $a$ and $b$.

$2.$ (added) If $x\ne 1$, then the equation can be manipulated to $y\ln(x-1)=\ln x+\ln(x-1)$. We recognize $y\ln(x-1)$ as the logarithm of $(x-1)^y$. So we can rewrite our equation as $(x-1)^y=x(x-1)$, which, since $x\ne 1$, can be simplified to $(x-1)^{y-1}=x$. It is likely that the solution can be written in terms of the Lambert $W$-function. A solution in terms of elementary functions seems highly unlikely.

  • $\begingroup$ yes, exactly. Thank you. $\endgroup$ – TomDavies92 Nov 12 '12 at 16:20

Yes, this is wrong, because you can use the property of ln, only when $ln \left ( \frac{a}{b} \right )=ln(a) - ln(b)$ Try to use the inverse of ln, we know from lections that it is the exponential function e.


This function is a really interesting one, I think. I've been looking at a similar one for a while: $y=\frac{\ln (x+1)}{\ln x}$. In fact, I think that my function is just yours but slid up one unit on both axes. I'm fairly sure that the inverses of your function can't be expressed with a finite number of elementary functions, or even a finite number of a wide variety of functions like the Lambert W function or the error function. I wasn't aware of your question, but I asked about my version of the function, and it got a bit more attention than this did. Maybe you can get something out of the answers that I got.

  • $\begingroup$ It is trivial to show that your function is the same as the OPs under the translation $x \mapsto x+1$, $y\mapsto y-1$. Of course, terms and conditions apply at points of discontinuity, etc. $\endgroup$ – Emily Dec 21 '12 at 19:13

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