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.
On
$$6x-3x^2=k^2$$
$$3x^2-6x+k^2=0$$
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 $k=\sqrt3$.
$$\log_k{(6x-3x^2)}=2$$ $$6x-3x^2=k^2$$ $$3x^2-6x+k^2=0$$ $$x=1\pm\sqrt{1-\frac{k^2}3}$$ 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 $x=1$.