2^(x)=16

The set A={x:x epsilonR,x^(2)=16 and 2x=16} is equal to

The set A={x:x epsilonR,x^(2)=16 and 2x=16} is equal to

Length of latus-rectum of Hyperbola y^(2)-16x^(2)=16 is -

Solve for x:5^(2x)-16.5^(x)-225=0

The roots of the equation (2-x)^(2)=16 are

The co-ordinates of the foci of 16y^(2) -x^(2) =16 are

The co-ordinates of the foci of 16y^(2) -x^(2) =16 are

((x)/(3)-(2)/(5))/((3)/(4)-2x)=(16)/(15)

If 2^(4)x4^(2)=16^(x), then find the value of x

Locus of centres of the circles which touch the two circles x^(2)+y^(2)=16 and x^(2)+y^(2)=16x externally is