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I was trying to calculate the inverse function ($f^{-1}$) of the following function: $$f(x)=1+x\cdot\ln(x)-\ln(x)^2 $$

It's a bijective function so $f^{-1}$ exists.

I tried doing the following:$$y=1+x\cdot\ln(x)-ln(x)^2 $$ $$e^y=e\cdot x^{x-2} $$

Then i got stuck, any hint would be helpfull. Thanks

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    There's very little you can do here. Even a simpler equation like $x^x =3$ requires Lambert W functions. Without inventing your own functions (which is not very satisfying) I think you're stuck permanently. – 2017-01-31
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    The title and the body do not agree. Which one is it? – 2017-01-31
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    I edited it, and yes i looked on the internet for solution but i think it is more complicated than i thought it was – 2017-01-31

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First thing here is the given function is not one-one. As $$f(1)=1=f(2).$$ Hence for this function inverse does not exist. But, like we do for trigonometric functions, we can restrict it to a smaller domain in which it is one-one and hence calculate its inverse in that domain.

But here the function is so complicated that we may not be able to write its inverse explicitly, like we don't know what is inverse of $cos(x)$ explicitly only we write it as $\arccos(x)$ and we can graph it.

Like we know the graph of the inverse function of $y=f(x)$ is obtained by reflecting it through $y=x$ line which is implicitly $x=f(y)$ so here it will be $$x=1+(y-2)\ln y$$ and you can plot it by help of any graphing calculator(one is Geogebra)

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    $f(2)=1,905...$ the inverse does exist. – 2017-02-01
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    is it $$f(2)=1+2\ln 2-\ln 4=1+2\ln2-2\ln2=1+0=1$$ incorrect?? – 2017-02-01
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    No you read it wrong it's $1+2\ln(2)-(ln(2))^2$ – 2017-02-01
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    Ok!! But you have written it is $\ln(x)^2\neq(\ln x)^2$. Then how you got $$e^y=e\cdot x^{x-2}??$$ – 2017-02-01
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    well, $e^{-\ln(x)^2}=e^{-2\ln(x)}=e^{\ln(x^{-2})}=x^{-2}$ – 2017-02-01
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    yeah!! that I'm saying. It is $\ln x^2=2\ln x$ not $(\ln x)^2=2\ln x$ – 2017-02-01
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52861/discussion-between-arun-badajena-and-hamza-boulahia). – 2017-02-01
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    still downvote?? – 2017-02-01
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    downvote = " the answer is not useful" and for the record yes it is not useful. – 2017-02-01