Respuesta :

konrad509
[tex]D:5x-2>0 \wedge x>0 \wedge x-1>0\\ D:5x>2 \wedge x>0 \wedge x>1\\ D: x>\frac{2}{5} \wedge x>1\\ D:x>1\\ \log_2(5x-2)-\log_2x-\log_2(x-1)=2\\ \log_2\frac{5x-2}{x(x-1)}=\log_24\\ \frac{5x-2}{x(x-1)}=4\\ 4x(x-1)=5x-2\\ 4x^2-4x=5x-2\\ 4x^2-9x+2=0\\ 4x^2-x-8x+2=0\\ x(4x-1)-2(4x-1)=0\\ (x-2)(4x-1)=0\\ x=2 \vee x=\frac{1}{4}\\ \frac{1}{4}\not \in D \Rightarrow \boxed{x=2} [/tex]

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