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I want so solve a problem:

for what values of a>0 does the equation a^x=x have solutions?

I have started to write it like a^logx=x. But its tricky this one. Does anyone have an idea of how to find the solutions?

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    why don't you try assessing the first derivative? :) – 2017-02-24
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    Is it clear to you that there must be a constant $c>1$ such that there are solutions exactly when $0 – 2017-02-24
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    yes thank u i will start with that! @llis – 2017-02-24
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    this is all i know about the funktion @HenningMakholm – 2017-02-24
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    @Fanny: When you say "this" do you mean what I just wrote? – 2017-02-24
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    @HenningMakholm i mean to me its not clear that there is such a boundary – 2017-02-24
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    @HenningMakholm or I guess It must be if the problem is to find that constant value c – 2017-02-24
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    @Fanny: Sonnhard's answer shows how to do that (it is somewhat hard to read, but the actual calculations are straightforward). – 2017-02-24
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    @HenningMakholm okej! im not so great with logarithms so i guess best to get to some simpler examples first – 2017-02-24

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i would write your equation like follows: $$\ln(a)=\frac{\ln(x)}{x}$$ and now we define $$f(x)=\frac{\ln(x)}{x}$$ for $$x>0$$ we get $$f'(x)=\frac{1-\ln(x)}{x^2}$$ we find $$f'(x)=0$$ if $$x=e$$ and $$f(e)=\frac{1}{e}$$ and $$f(e)$$ is the Maximum of f(x). Nowe we get if $$-\infty\frac{1}{e}$$ our equation has no solution. Note that $$y=f(x)=\frac{\ln(x)}{x}$$