[tex]4^{2-x} [/tex] and [tex]7^{3-x}[/tex] are positive for all x∈R, so we can logarithm both sides. There will be:
[tex]4^{2-x}=7^{3-x}\quad|\ln(\dots)\\\\\ln4^{2-x}=\ln7^{3-x}\\\\(2-x)\ln4=(3-x)\ln7\\\\2\ln4-x\ln4=3\ln7-x\ln7\\\\x\ln7-x\ln4=3\ln7-2\ln4\\\\x(\ln7-\ln4)=3\ln7-2\ln4\quad|:(\ln7-\ln4)\\\\\\\boxed{x=\dfrac{3\ln7-2\ln4}{\ln7-\ln4}}[/tex]
This is exact form of solution. Now, we can use tables of logarithms or calculator and calculate approximate form. We get:
[tex]\boxed{x\approx5,477}[/tex]
