Solve $2^{\left(\log_2\sqrt{x^2-6x+9}\right)} = 3^{\log_\sqrt{x}(x-1)}$ for real $x$

Question

Solve $$ 2^{\left(\log_2\sqrt{x^2-6x+9}\right)} = 3^{\log_\sqrt{x}(x-1)} $$ for any real $x$.

I am unable to solve it completely, the farthest I went (for $x \in (0,3)$) was $\log(x)\cdot \log(3-x) = \log(x-1) \cdot \log(9)$.

Answer 1

Use $a^{\log_a x}=x$, then we get $$\sqrt{x^2-6x+9}=3^{\log_{\sqrt{x}} (x-1)}\implies |x-3|= 3^{\log_{\sqrt{x}}(x-1)}.$$ The root can be found by ispection only and it is $x=2.$

Answer 2

If you plot the function $$f(x)=\sqrt{x^2-6 x+9}-3^{\frac{\log (x-1)}{\log \left(\sqrt{x}\right)}}$$ you can notice tht the second root is close to $x=11$.

So, make a Taylor expansion around $x=9$ to have nice numbers to get $$f(x)=-2+\frac{12 \log (2)}{9 \log (3)}(x-9)+O\left((x-9)^2\right)$$ Ignoring the higher order terms, an approximation is $$x=9+\frac{3\log(3)}{2\log(2)}\approx 11.3774$$ while the "exact" solution could be obtained very fast starting at this value. $$\left( \begin{array}{cc} n & x_n \\ 0 & 11.377443751081734272 \\ 1 & 11.272854680548284363 \\ 2 & 11.272737325369640493 \\ 3 & 11.272737325218626553 \end{array} \right)$$

Answer 3

We have:

$\begin{align*} 2^{\log_2 \sqrt{x^2 -6 x + 9}} &= 3^{\log_{\sqrt{x}}(x-1)} \\ \sqrt{(x - 3)^2} &= 3^{\log_3 (x - 1) / \log_3 \sqrt{x}} \\ \lvert x - 3 \rvert &= (x - 1)^{2/ \log_3 x} \\ \lvert x - 3 \rvert^{\log_3 x} &= (x - 1)^2 \end{align*}$

This is as far as simplification of the mess takes you. Now you have to switch to analyze cases. First off, $x > 0$, as otherwise the logarithm makes no sense. Need to consider $0 < x \le 1$, $1< x \le 3$, $x > 3$ separately. Perhaps plot the functions to see how they behave to see how to prove there is no/one/more zeros in each range.