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  4. If log2(2x-1 + 6) + log2(4x-1 ) = 5 , then x =

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Q. If $\log_2(2^{x-1} + 6) + \log_2(4^{x-1} ) = 5 ,$ then $x = $

COMEDKCOMEDK 2009Probability - Part 2

Solution:

We have, $\log_2(2^{x-1} + 6) + \log_2(4^{x-1}) = 5 \, \log_2[(2^{x-1} + 6) 4^{x-1}] = 5$ .
$\Rightarrow (2^{x-1} + 6) 4^{x-1} = 2s \Rightarrow (2^{x-1} + 6) (2^{x-1} )^2 = 32$ $\Rightarrow (t + 6) t^2 = 32 (put 2x-l = t)$ $\Rightarrow t^3 + 6t^2 - 32 = 0 \Rightarrow t = 2 \Rightarrow 2^{x-1} = 2$ $\Rightarrow x-1=1 \Rightarrow x=2$