Question

The number of solution of 10^(2/x) + 25^(1/x) = 65/8(50^(1/x))

a)" "1

b)" "2

c)" "3

d)" "oo

Solution

10^(2/x) + 25^(1/x) = 65/8(50^(1/x))

=> (2xx5)^(2/x) + (5xx5)^(1/x) = 65/8(2xx5xx5)^(1/x)

=> (2^(2/x))xx(5^(2/x)) + 5^(2/x) = 65/8(2^(1/x))xx(5^(2/x))

=> (5^(2/x)) [(2^(2/x)) + 1] = 65/8(2^(1/x))xx(5^(2/x))

So now, (5^(2/x)) = 0, this is not possible

and (2^(2/x)) + 1 = 65/8(2^(1/x))

Putting, (2^(1/x)) = z

Then, z^2 + 1 = 65/8z

=> 8z^2 + 8 = 65z

=> 8z^2 - 65z + 8 = 0

=> 8z^2 - 64z - z + 8 = 0

=> 8z(z-8) - 1(z-8) = 0

=> (8z-1)(z-8) = 0

So, z = 1/8 and 8

=> 2^(1/x) = 1/8 and 8

This will give, x = -1/3 and 1/3