Let me see if I guessed correctly what you wrote:
$$4\log_{x/2}\sqrt x+2\log_{4x}x^2=3\log_{2x}x^3\iff2\log_{x/2}x+4\log_{4x}x=9\log_{2x}x\implies$$
$$2\frac{\log x}{\log\frac x2}+4\frac{\log x}{\log 4x}=9\frac{\log x}{\log 2x}$$
where $\;log\;$ above can be at any base you want (though in higher mathematics it is usually taken to be $\;\log_e\;$)
Now you could assume $\;x\neq 1\;$ (otherwise the exercise is very boring) and thus divide through the whole last equation by $\;\log x\;$ , and then use other properties of logarithms, say like $\;\log AB=\log A+\log B\;$ , or $\;\log\frac AB=\log A-\log B\;$ , etc.
Also make sure you understand and can justify the first steps shown above.
2 log(x^2) base(4x)
? I don't even know if it's a allowed to have a non-constant variable as the base, but you may be able to use $\log_b(a) = \frac{\ln(a)}{\ln(b)}$. $\endgroup$