Question

Срочно, помогите решить

Answer 1

$\displaystyle 1)log_3(27)-log_1_0(0,01)+log_0_,_6(0,36)+log_2(1)=log_3(3^3)-log_1_0(10^-^2)+\\ \\ log_{\frac{3}{5} }((\frac{3}{5})^2)+0=3-(-2)+2=3+2+2=7 \\\\2)\sqrt[5]{2^1^0*7^5}=2^2&*7=4*7=28\\\\3)log_2(x-2)<4,x>2\\x-2<2^4\\x-2<16\\x<16+2\\x<18\\x \in (2,18)\\\\4)log_{\frac{1}{4} }(x+3)<2,x>-3\\\\x+3> (\frac{1}{4})^2\\\\ x+3>\frac{1}{16}\\\\x>\frac{1}{16}-3\\\\x>-\frac{47}{16}\\\\x \in (-\frac{47}{16},+ \infty)$