Verified answer

[tex]\displaystyle\bf\\3)\\\\2^{x} \cdot 5^{\frac{1}{x} } > 10\\\\\log_{2} \Big(2^{x} \cdot 5^{\frac{1}{x} } \Big) > \log_{2} 10 \ \ \ , \ \ \ x\neq 0\\\\\\\log_{2} 2^{x} +\log_{2} 5^{\frac{1}{x} } > \log_{2} (2\cdot 5)\\\\\\x+\frac{1}{x} \log_{2} 5 > \log_{2} +\log_{2} 5\\\\\\x+\frac{1}{x} \log_{2} 5-(1+\log_{2} 5) > 0\\\\\\\frac{x^{2} -x\cdot(1+\log_{2} 5)+\log_{2} 5}{x} > 0\\\\\\\frac{x^{2} -x-x\log_{2} 5+\log_{2} 5}{x} > 0\\\\\\\frac{(x^{2} -x\log_{2} 5)-(x-\log_{2} 5)}{x} > 0[/tex]

[tex]\displaystyle\bf\\\frac{x\cdot(x -\log_{2} 5)-(x-\log_{2} 5)}{x} > 0\\\\\\\frac{(x -\log_{2} 5)\cdot(x-1)}{x} > 0\\\\\\- - - - (0)+ + + + (1)- - - - (\log_{2} 5) + + + + \\\\\\Otvet \ : \ x\in(0 \ ; \ 1)\cup(\log_{2} 5 \ ; \ +\infty)[/tex]