(1)/(4-x^(2))+log_(10)(x^(3)-x)

If (1 + 3 + 5 + .... " upto n terms ")/(4 + 7 + 10 + ... " upto n terms") = (20)/(7 " log"_(10)x) and n = log_(10)x + log_(10) x^((1)/(2)) + log_(10) x^((1)/(4)) + log_(10) x^((1)/(8)) + ... + oo , then x is equal to

If (1 + 3 + 5 + .... " upto n terms ")/(4 + 7 + 10 + ... " upto n terms") = (20)/(7 " log"_(10)x) and n = log_(10)x + log_(10) x^((1)/(2)) + log_(10) x^((1)/(4)) + log_(10) x^((1)/(8)) + ... + oo , then x is equal to

Solve for x, (a) (log_(10)(x-3))/(log_(10)(x^(2)-21))=(1)/(2),(b)log(log x)+log(log x^(3)-2)=0; where base of log is 10 everywhere.

If 1, log_(10)(4^(x)-2) and log_(10)(4^(x)+(18)/(5)) are in arithmetic progression for a real number x, then the value of the determinant |(2(x-(1)/(2)), x-1,x^(2)),(1,0,x),(x,1,0)| is equal to:

The inverse of f(x)=(10^(x)-10^(-x))/(10^(x)+10^(-x)) is A). (1)/(2)log_(10)((1+x)/(1-x)) , B). log_(10)(2-x) , C). (1)/(2)log_(10)(2-1) , D). (1)/(4)log_(10)((2x)/(2-x))

Solve |x-1|^((log_(10)x)^(2)-log_(10)x^(2))=|x-1|^(3)

Solve: |x-1|^((log_(10)x)^(2)-log_(10)x^(2))=|x-1|^(3)

The solution set of the equation log_(10)(3x^(2)+12x+19)log_(10)(3x+4)=1 is

If x_(1),x_(2)&x_(3) are the three real solutions of the equation x^(log_(10)^(2)x+log_(10)x^(3)+3)=(2)/(((1)/(sqrt(x+1-1))-(1)/(sqrt(x+1+1)))) where x_(1)>x_(2)>x_(3), then

If f(x)=cos^(-1)((2-|x|)/(4))+[log_(10)(3-x)]^(-1) , then its domain is