What is $x$ satisfying the condition?

Question

What is $x$ such that $3^{\log (x)}-2^{\log(x)}=2^{\log (x+1)}-3^{\log (x-1)}$?

Wolfram says $x=2.47\ldots$ but I couldn't get it manually.

Thanks in advance for any help!

Answer 1

We can obtain decent approximation of the solution.

Let $x=y+1$ and consider that you look for the zero of function $$f(y)=\Big[3^{\log (y+1)}-2^{\log(y+1)}\Big]-\Big[2^{\log (y+2)}-3^{\log (y)}\Big]$$ Using $y_0=1$, perform one single iteration of Newton-like methods of order $n$ $$\left( \begin{array}{ccc} n & y_n & \text{method} \\ 2 & 1.505656785 & \text{Newton} \\ 3 & 1.475784581 & \text{Halley} \\ 4 & 1.480463468 & \text{Householder} \\ 5 & 1.479522210 & \text{no name} \\ 6 & 1.479745762 & \text{no name} \\ \cdots & \cdots & \\ \infty & 1.479698894 & \text{no name} \end{array} \right)$$