if equation `log_4(2log_3(6+log_2(x^2+kx+9)))`=1 has exactly two distinct real solutions, then least positive integral value of k, is

If 5x^2-2kx+1 < 0 has exactly one integral solution which is 1 then sum of all positive integral values of k.

If the equation x ^(4)+kx ^(2) +k=0 has exactly two distinct real roots, then the smallest integral value of |k| is:

If the equation ln (x^(2) +5x ) -ln (x+a +3)=0 has exactly one solution for x, then possible integral value of a is:

If the equation ln (x^(2) +5x ) -ln (x+a +3)=0 has exactly one solution for x, then possible integral value of a is:

The equation (log)_2(2x^2)+(log)_2xdotx^((log)_x((log)_2x+1))+1/2log4 2(x^4)+2^(-3(log)_(1/2)((log)_2x))=1 has exactly one real solution two real solutions three real solutions (d) no solution

The equation (log)_2(2x^2)+(log)_2xdotx^((log)_x((log)_2x+1))+1/2log_4^2(x^4)+2^(-3(log)_(1/2)((log)_2x))=1 (a)has exactly one real solution (b)two real solutions (c)three real solutions (d) no solution

If the equation x^(log_(a)x^(2))=(x^(k-2))/a^(k),a ne 0 has exactly one solution for x, then the value of k is/are

If x^2-4x+log_(1/2)a=0 does not have two distinct real roots, then maximum value of a is

The equation (log)_(x+1)(x- .5)=(log)_(x-0. 5)(x+1) has (A) two real solutions (B) no prime solution (C) one integral solution (D) no irrational solution

If x^(4) + 2kx^(3) + x^(2) + 2kx + 1 = 0 has exactly tow distinct positive and two distinct negative roots, then find the possible real values of k.