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?
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IV_
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TomDavies92
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Just out of curiosity: where did you come across this problem? – Thomas Nov 12 '12 at 16:42
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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.
André Nicolas
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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.
Lullaby
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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.
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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. – Emily Dec 21 '12 at 19:13
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$$y=\frac{\ln(x)}{\ln(x-1)}+1$$
For real $a,b$, there is the rule $\ln(a\cdot b)=\ln(a)+\ln(b)$, not your simplification.
Your function is an elementary function. It's a binary algebraic function of two algebraically independent monomials. It doesn't have an elementary inverse therefore.
$$y-1=\frac{\ln(x)}{\ln(x-1)}$$
$$1=\frac{\ln(x)}{\ln(x-1)(y-1)}$$
$$\ln(x-1)=\frac{\ln(x)}{(y-1)}$$
$$x-1=e^{\frac{\ln(x)}{(y-1)}}$$
Lambert W cannot be applied here.
$$x-1=x^{\frac{1}{y-1}}$$
For rational $y\neq 1,2$, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For rational $y\neq 1,2$, the equation is related to a trinomial equation.
For real or complex $y\neq 1,2$, the equation is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.
$\ $
IV_
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