log2002 base n -> log2002 base n
find the value of
log22002 + log32002 + log42002 + log52002 - log102002 - log112002 - log122002 - log132002 - log142002
Option
a) 0
b) 1
c) -1
d) 2002log2002

• 
  log a base b
  logba=loga/logb
  so
  log2 2002+log3 2002 +...-(log10 2002+log11 2002+...)
  log2002/log2 + log2002/log3+... -(log2002/log10 + log2002/log11 +...)
  log2002 ((1/log2 + 1/log3+...)-(1/log10 + 1/log11 +...))
  log2002 (((log2)^-1 + (log3)^-1 + ...) - ((log10)^-1 + (log11)^-1 + ...))
  log2002 ((-log1 - log2 - ...)-(-log10 -log11 -....))
  log2002 ((log10 + log 11 + ...)-(log2+log3+...))
  log2002 (log ((10*11*12*13*14)/(2*3*4*5))
  log2002 (log2002)
  log2002/(log2002)^-1
  -log2002/log2002
  -1 ANS
  
• 10 years agoHelpfull: Yes(34) No(10)
• log a base b means loga/logb so if we do this log(2*3*4*5/10*11*12*13*14) all log 2002 cancelled out so log(01/11*13*14) is negative so ans is -1
• 10 years agoHelpfull: Yes(2) No(2)
• Ans -1 is correct.
• 9 years agoHelpfull: Yes(1) No(0)
• on solving you finally get (log 2002)(log 2002)=(log 2002)^2 which is somewhere between 9 and 15 which is not found in any of the options.Hence either the question is wrong or the options are inappropriate......
• 10 years agoHelpfull: Yes(0) No(0)