If `(log)_(0. 3)(x-1)<(log)_(0. 09)(x-1),` then `x` lies in the interval `(2,oo)` (b) `(1,2)` `(-2,-1)` (d) None of these

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))/(x^(2)-3x-18)) is

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))/(sqrt(x^(2)+2x+8))" is"

(log)_(0.3)(x-1)<(log)_(0.09)(x-1)thenx in the lies in the interval- a.(2,oo) b.(1,2) c.(-2,-1) d.(-oo,2)

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)]

Solve :log_(0.3)(x^(2)-x+1)>0

Find the values of x satisfying the inequalities: log_(0.1)(4x^(2)-1)>log_(0.1)3x

Solve the following inequalities (i) log_(5)(3x-1) lt 1 (ii) (log_(.5)x)^(2)+log_(.5)x-2 le 0 (iii) log_(3)(x+1)+log_(3)(x+7) ge 3 (iv)log_(1//2)log_(3)(x^(2)+5)+1 le 0