If `log_(3/2)((2x-8)/(x-2))<0` then set of all possible values of `x` is :

Solve for x:(a) log_(0.3)(x^(2)+8) gt log_(0.3)(9x) , b) log_(7)( (2x-6)/(2x-1)) gt 0

If log_(2)(3^(2x-2)+7)=2+log_(2)(3^(x-1)+1) then x=

Find x if it is given by log_(8)((8)/(x^(2)))=3(log_(8)x)^(2)

The equation (log_(8)((8)/(x^(2))))/((log_(8)x)^(2))=3 has

Solve the following equations : (i) log_(x)(4x-3)=2 (ii) log_2)(x-1)+log_(2)(x-3)=3 (iii) log_(2)(log_(8)(x^(2)-1))=0 (iv) 4^(log_(2)x)-2x-3=0

If log_(3x+5)(ax^(2)+8x+2)>2 then x lies in the interval

Equation 2log_(8)(2x)+log_(8)(x^(2)+1-2x)=(4)/(3) has

If log_(2)(log_(8)x)=log_(8)(log_(2)x), find the value of (log_(2)x)^(2)

The expression: (((x^(2)+3x+2)/(x+2))+3x-(x(x^(3)+1))/((x+1)(x^(2)+1))-log_(2)8)/((x-1)(log_(2)3)(log_(3)4)(log_(4)5)(log_(5)2)) reduces to

If log_(2) 6 + (1)/(2x ) =log_(2) (2^(1//x) + 8) , then the values of x are