if `log_(0.3)(x-1)`<`log_(0.09)(x-1)` the `x` will lie in the interval

Find the value of x such that log_(0.3)(x-1) = log_(0.09)(x-1) .

The domain of the function f(x)=sqrt((log_(0.3)(x-1))/(x^(2)-3x-18)) is

The domain of the function f(x)=sqrt(-log_(0.3) (x-1))/(sqrt(-x^(2)+2x+8))" is"

The domain of the function : f(x)=(sqrt(-log_(0.3)(x-1)))/(sqrt(-x^(2)+2x+8)) is :

The domain of the function f(x) = sqrt(-log_(0.3)(x-1))/sqrt(x^2 + 2x + 8) is

The domain of the function f(x)=sqrt(-log_(0.3) (x-1))/(sqrt(x^(2)+2x+8))" is"

The domain of the function f given by f(x)=sqrt(log_(0.3)((e^(x)-1)/(e^(x)-tan x))) contains (A) (0,(pi)/(4)](B)[(pi)/(4),(pi)/(2))(C)[-(3 pi)/(4),-(pi)/(2)) (D) ((pi)/(2),(5 pi)/(4)]

The value of x, satisfying the inequality log_(0.3)(x^(2)+8)>log_(0.3)9x, lies in