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Question

The value of x satisfying log16x+logx16=log512x+logx512 is/are
  1. 164
  2. 116
  3. 64
  4. 16


Solution

The correct options are
A 164
C 64
Given 
log16x+logx16=log512x+logx51214log2x+4logx2=19log2x+9logx2
Assuming log2x=t, we get
t4+4t=t9+9tt4t9=9t4t5t36=5tt2=36t=±6log2x=±6x=164,64

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