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How can I prove that this has only one solution for 0< x <2 when f(x)=2?

$$f(x)= \log_{k} (6x-3x^2)$$

When I try to solve this equation I get $$k^2=-3x(x-2)$$

But then I'm stuck as it feels like it leads to an infinity of results.

I sense that calculus should be involved but I don't know how or why.

Thanks.

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    $\begingroup$ What do you mean by solutions? Roots of $f$? $\endgroup$ – Eevee Trainer May 11 at 9:47
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    $\begingroup$ I expect that the claim is one solution for a specific $k$ and not that you can determine both $x$ and $k$ from that. $\endgroup$ – badjohn May 11 at 9:48
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    $\begingroup$ You mean $f(x)=2$?But $f(x)=f(2-x)$. $\endgroup$ – Oolong milk tea May 11 at 9:51
  • $\begingroup$ Yes, My bad, I forgot the essential: for f(x)=2, indeed! $\endgroup$ – Bachir Messaouri May 11 at 9:53
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$$\log_k{(6x-3x^2)}=2$$ $$6x-3x^2=k^2$$ $$3x^2-6x+k^2=0$$ $$x=1\pm\sqrt{1-\frac{k^2}3}$$ Now assuming that $0\lt k\lt\sqrt{3}$ (and $k\ne1$) there must be two solutions for $x$ in this range. For your claim to be true, you need that $k=\sqrt{3}$ such that the only possible solution is when $x=1$.

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  • $\begingroup$ Thank you! It’s cristal clear! $\endgroup$ – Bachir Messaouri May 11 at 10:12
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$$6x-3x^2=k^2$$

$$3x^2-6x+k^2=0$$

Let $g(x) = 3x^2-6x+k^2=3(x-1)^2+k^2-3$, $g$ is a convex quadratic function that is symmetric about $x=1$. Note that this is true for $6x-3x^2$ as well.

Hence if there is any $x \ne 1$ that is a root in $[0,1]$, there will be at least two roots. If there is a unique root, it has to be attained at $x=1$ which restrict $-3+k^2=0$ and $k=\sqrt3$.

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  • $\begingroup$ Thank you very much! $\endgroup$ – Bachir Messaouri May 11 at 10:12

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