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# 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) AskIITians Expert Hari Shankar IITD
17 Points
12 years ago

Hi

By the rules of logs, logabc = c logab and logab c = (1/b) logac

Applying these rules, we get

log2x4 = 4 log2x

Also, log4(2/x)1/2 = (1/2) log4(2/x)

= (1/2) log22(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