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) .

If log_(0.04)(x-1)>=log_(0.2)(x-1) then x belongs to the interval

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 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

For what values of x,log_(0.3)(x^(2)+8)>log_(0.3)(9x)