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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 8y 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!!!!
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