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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)
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)
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, 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
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
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