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by manpsingh87 » Sun Jun 12, 2011 8:09 am
vinayreguri wrote:2^(2x+8)+1=32*2^x
A. 4
B. 2
C. 0
D. -2
E -4

[spoiler]OA : E[/spoiler]
Best is to substitute each AO and find it, but if the choices are not given, how?
2^(2x+8)+1=32*2^x ;
2^2x*256+1=32*2^x;
let 2^x=y;-----------1)
256y^2-32y+1=0;
(16y-1)=0;
y=1/16;
from 1) we have;
2^x=1/16;
2^x=2^-4;
hence x=-4, therefore E
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by Anurag@Gurome » Sun Jun 12, 2011 8:53 am
vinayreguri wrote:2^(2x+8)+1=32*2^x
A. 4
B. 2
C. 0
D. -2
E -4
Say, 2^x = p --> 32*(2^x) = 32p
And, 2^(2x + 8) = (2^2x)*(2^8) = 256*(2^2x) = 256p²

Hence, 256p² + 1 = 32p
---> 256p² - 32p + 1 = 0
---> (16p - 1)² = 0
---> p = 1/16

Therefore, 2^x = 1/16 --> x = -4

The correct answer is E.
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by cans » Tue Jun 14, 2011 4:07 am
let 2^x=t
256t^2 -32t + 1 =0
(16t -1)^2 = 0
t=1/16
x=-4
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