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$\log _4 2-\log _g 2+\log _{16} 2-...............=$

\[\begin {array} {1 1} (1)\;e^2 & \quad (2)\;\log _e^2 \\ (3)\;1+\log _e 3 & \quad (4)\;1- \log _e2 \end {array}\]

1 Answer

$ (4)\;1- \log _e2 $
answered Nov 7, 2013 by pady_1
 
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