# Function has only one solution

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.

• What do you mean by solutions? Roots of $f$? – Eevee Trainer May 11 at 9:47
• I expect that the claim is one solution for a specific $k$ and not that you can determine both $x$ and $k$ from that. – badjohn May 11 at 9:48
• You mean $f(x)=2$?But $f(x)=f(2-x)$. – Oolong milk tea May 11 at 9:51
• Yes, My bad, I forgot the essential: for f(x)=2, indeed! – Bachir Messaouri May 11 at 9:53

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

• Thank you! It’s cristal clear! – Bachir Messaouri May 11 at 10:12

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

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

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 $$k=\sqrt3$$.

• Thank you very much! – Bachir Messaouri May 11 at 10:12