Find the value of ' x ' , if (\frac{1}{2^{\log_x4}})( {1}/{2^{\log _{x} 16}})( {1}/{2^{\log _{x} 256}}) . . . \infty=2

NIMCET 2009 Question Paper with solutions

Question : 43 of 120

Marks: +1, -0

Find the value of ' x ' , if

(\;{1}/{2^{\log_x4}})(\;{1}/{2^{\log _{x} 16}})(\;{1}/{2^{\log _{x} 256}}) . . . ∞=2

Solution:

From given relation, we get.

2^{\log _{x} 4+\log _{x} 16+\log _{x} 256+. . . x}=\;{1}/{2}

⇒ 2^{\;{1}/{\log x}[\log 4+\log 4^{2}+\log 4^{4}+. . . ∞)}=\;{1}/{2}

2^{{\log 4}/{\log x}(1+2+4+.......∞)}=\;{1}/{2}

⇒ \;\; 2^{\;{\log 4}/{\log x}}\{\table 1 ; {1}-2\}=\;{1}/{2}

⇒ \;\; 2^{-\log _{x}} 4=2^{-1} ⇒ \log _{x} 4=1 ⇒ x=4

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