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Vaibhav Mathur Grade: 12
        

 

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)



7 years ago

Answers : (1)

AskIITians Expert Hari Shankar IITD
17 Points
										

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


 


 


 

7 years ago
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