I am still confused with the problem Simply Exponential \[\quad \quad \quad \quad { x }^{ x }=x\\ \Longrightarrow x={ x }^{ 1/x }\\ \Longrightarrow { x }^{ x }={ x }^{ 1/x }\\ \Longrightarrow x\ln { x } =\frac { 1 }{ x } \ln { x } \\ \Longrightarrow \ln { x } (\frac { { x }^{ 2 }-1 }{ x } )=0\ \\ \Longrightarrow x=\pm 1(Correct \quad answer)\] Now,since we have included a step \(x\ln { x } =\frac { 1 }{ x } \ln { x }\) in our solution,according to the domain of the logarithmic function, x should be positive,thus omitting -1.Then why -1(as we cannot take log of a negative number)?

**Then, I am confused that the given expression(actually an equation) is a function or not.** I mean, if it is given that \(f\left( x \right) ={ x }^{ x }-x\) and then we are asked to find the roots or zeroes of the function, then the answer will be only +1 or still be both+1 & -1?(I think that it should be only +1 as the domain of \(x^{x}\) is \(x>0\) )

Also, I don't know why, but is the second step correct and safe?

Please clear my doubts,and please provide any kind of guidance wherever you think is needed. Thanks

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TopNewestYou are right that, if we just want to work with continuous functions over the field of real numbers, the domain of both \(x^{x}\) and \(\ln(x)\) is \(x \gt 0.\) So the step where we take the natural log of both sides can only be done under the condition that \(x \gt 0,\) and hence with this process only the solution \(x = 1\) is valid.

However, the function \(f(x) = x^{x}\) is also purely real, (i.e., has an imaginary component of \(0*i\)), whenever \(x\) is a negative integer, and since \((-1)^{-1} = \dfrac{1}{(-1)^{1}} = -1\) we also have that \(x = -1\) is a solution to the equation \(x^{x} = x.\)

So while \(g(x) = x^{x} - x\) is a continuous function only for \(x \gt 0,\) it does still have a "zero" at \(x = -1.\)

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Sir, if -1 is a zero of the

function,then what will be it's(I mean the function's) domain? (As the zero sholud lie in it's domain).Will it's domain be (all positive real numbers + negative integers)? Also, according to you, -1 is not a solution which we get mathematically(I mean analytically) and it is just by random hit and trial, so if in a subjective paper, we have to conclude -1(it is a bit funny and pointless, but just...),what's the best way to do it?

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I was wondering that myself. WolframAlpha lists the domain only as \(x \gt 0,\) and yet it also states that \(x = -1\) is a root, which at first glance seems like a contradiction; how can a function have a root outside of its domain? The only way I can resolve this is that as a continuous function its domain is \(x \gt 0,\) and that \(x = 1\) is technically the only root, but it also happens to be the case that \(g(-1) = 0.\) I'm not sure if we should really call \(x = -1\) a root, (or zero), but nevertheless \(x = -1\) is a valid solution to the equation \(x^{x} = x.\) So for now I would still consider the domain as just \(x \gt 0,\) and that the solution \(x = -1\) is, as you say, a hit and trial "anomaly".

When dealing with powers of negative numbers it makes more sense to work in the field of complex numbers. That way we can expand the domain and such anomalies then become "analytic" solutions.

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Of course, for a function to be completely well defined, the domain needs to be stated. In the event that the domain isn't stated, it is extremely open to interpretation. E.g. What are the zeros of \( f(x) = x^2 + 1 \)? Those who say no zeros (working in the domain of real numbers) are equally correct as those who say zero of \(i \) and \( -i \) (working in the domain of complex numbers).

In the event that the domain is over specified, then we simply ignore values where the function doesn't make sense. In this case, it means that we remove those negative non-integer values.

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See Ivan's solution, which avoids the numerous issues that you have raised.

Whenever we manipulate an equation, we have to be careful about the steps that we're taking, and check to see if we're introducing extraneous solutions, or losing actual solutions. For example, by taking logs, we need to assume that \( x > 0 \).

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@Janardhanan Sivaramakrishnan Sir & @Brian Charlesworth Sir please help.

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\(x \ln x = \frac{1}{x} \ln x\)

Let \(x=r e^{i\theta}\). Then, \(\ln x = \ln r + i\theta\) (This step makes it possible to have natural logarithms for negative or even complex numbers).

\((\ln r + i\theta)(r e^{i\theta} - r^{-1} e^{-i\theta})=0\)

Case 1 : \(\ln r = 0\) and \(\theta = 0\). This implies \(x=1\)

Case 2 : \(r e^{i\theta} = r^{-1} e^{-i\theta}\)

This means \(r^2 e^{2i\theta} = 1\)

Which in turn means that \(|r| =1\) and \(2\theta = 2n\pi\), or \(\theta = 0,\pi\)

This gives \(x=1\) again. But, we also find the second solution \(x=-1\).

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Thanks Sir for clearing the confusion.

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I think \( x = +1\) is the only correct solution because domain of our function is \( x > 0 \)

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It is not stated that the domain is \( x > 0 \). It asks for "distinct real solutions".

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If some function has domain \(x > 0\) ,then how could it have a negative solution?

Before finding the solution of some function, don't we first check its domain?

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If, however, we were asked for the number of roots of the function \(f(x) = x^{x} - x,\) then we would have be mindful of the domain. If we establish the domain as \(x \gt 0\) then there would be only one valid root, namely \(x = 1.\) If we were to also include the set of all negative numbers in the domain then \(x = -1\) would also be a valid root. The more "standard" choice of domain would be \(x \gt 0,\) though, since \(f(x)\) would then be continuous and differentiable over all the domain.

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Thanks for inserting the graph. Yes, analytically,only +1 is the solution or zero of the function(still, -1 is confusing as \(f(-1)=0\)),as mentioned by Brian Sir, but -1 is a valid solution of the equation \(x^{x}=x\) [if not considering it a function],which we get by hit & trial.

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From where do you get this question?

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BTW, Sir, can you please go through my note and comments and tell if -1 will be a zero of the function or not. Thanks

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