Find the no. of values of x belongs to R for which

64^(1/x) + 48^(1/x) = 80^(1/x)

plzz give the full soln...

Let 1/x = y.

The equation then becomes . Divide by 8^y on both sides to obtain 

. If y<0, then there are no solutions.

LHS is a strictly monotonic function in y. Hence there can exist at most one solution to the equation. y = 2 is a solution and hence the unique solution. (You can also use the Power Mean Inequality to arrive at this conclusion)

Hence x = 1/2 is the only solution.

Given:

  (64)^1/x +(48)^1/x=(80)^1/x

=>Tahe 16^1/x common on both side s.t.

(16X4)^1/x +  (16X3)^1/x =(16X5)^1/x

=>4^1/x + 3^1/x=5^1/x

=>By pythgerous triplet

 4^2 + 3^2=5^2

=>1/x=2

=>x=1/2=0.5

e1 i got the same answer

but the answer given is that there r 4 solutions!!!!