Răspuns :
[tex]\displaystyle\bf\\lg((x-1)^{10})<10lg(x)\\Conditii:\\lg(x)~~~~~~x>0\\lg(x-1)~~x-1>0~\implies~x>1\\Pe~ansamblu: \boxed{\bf~x>1}\\\\Rezolvare:\\\\lg((x-1)^{10})<10lg(x)~~~Exponentul~trece~in~fata.\\\\10lg(x-1)<10lg(x)~\Big|:10\\\\lg(x-1)<lg(x)\\\\lg(x-1)-lg(x)<0\\\\lg\left(\frac{x-1}{x}\right)<0\\\\\textbf{Functia logaritm este negativa daca argumentul este subunitar.}[/tex]
.
[tex]\displaystyle\bf\\0<\frac{x-1}{x}<1\\\\Conditia~~~0<\frac{x-1}{x}~~~nu~este~necesara~deoarece~avem~conditia:\\x>1\\\textbf{Pentru orice valoare a lui x mai mare decat 1, }\\\textbf{fractia nu poate fi negativa.}\\\\Rezolvam~inecuatia:\\\\\frac{x-1}{x}<1~~\Big|\cdot~x\\\\x-1<x\\\\x-1-x<0\\x-x<1\\0<1~~~(Adevarat)\\\\\textbf{Rezulta ca inegalitatea este adevarata pentru oricare }~x\in R\\\\Dar~avem~conditia:\\x>1\\\\\implies~\boxed{\bf~x\in(1,~\infty)}[/tex]
