Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

The root(s) of the equation (log 2 x 4 )^1/2 +4log 4 (2/x)^1/2=2 lie in the interval (a)(-200,-100) b. (-100,0) (c) (0,100) d. (100,200)

 

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)



Grade:12

1 Answers

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

 

 

 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free