The root(s) of the equation (log₂ x⁴)^1/2 +4log₄(2/x)^1/2=2 lie in the interval

(a)(-200,-100)

b. (-100,0)

(c) (0,100)

d. (100,200)

Hi

  By the rules of logs, log_ab^c = c log_ab and log_a^b c = (1/b) log_ac

Applying these rules, we get

log₂x⁴ = 4 log₂x

Also, log₄(2/x)^1/2 = (1/2) log₄(2/x)

= (1/2) log_2²(2/x) = (1/2)(1/2) log₂(2/x)

= (1/4) log₂(2/x) = (1/4)[log₂2 - log₂x] =  (1/4)(1-y)

[Where y = log₂(x)]

So the given equation becomes

(4 log(x))^1/2 + 4(1/4)(1-y) = 2

or, (4y)^1/2 + (1-y) = 2

Let y = u².

Now it becomes

2u + (1-u²) = 2

or u²-2u+1 = 0

or (u-1)² = 0

Therefore u = 1.

So, y = u² = 1 and hence, log₂(x) = 1.

So, the solution is x=2, which lies in the interval specified by option C