solve [logap(log(a/p)]/[logpa(log(p/a))] where logap is log base a p

• ans will be -logap
• 9 years agoHelpfull: Yes(6) No(0)
• since logap=loga+logp=logpa ,log(a/p)=loga-logp and log(p/a)=logp-loga;
  put the value in the given equation
  we get
  [(loga+logp)(loga-logp)]/[(logp+loga)(logp-loga)]
  which is equal to -1.
• 9 years agoHelpfull: Yes(3) No(8)
• Ans:(-logap)
• 9 years agoHelpfull: Yes(2) No(0)
• z= [log(base ap)(log(a/p)]/[log(base ap)(log(p/a))]
  => z= [log(base ap)(log(ap^-1))]/[log(base ap)(log(a^-1p))]
  => z= -1[log(base ap)(log(ap))] / -1[log(base ap)(log(ap))]
  => z= -1/-1=1 ...ans
• 8 years agoHelpfull: Yes(2) No(1)
• ans is -1 for sure
  
• 8 years agoHelpfull: Yes(1) No(1)
• answer will be -1
  
• 9 years agoHelpfull: Yes(0) No(1)
• Can you please explain the answer
• 9 years agoHelpfull: Yes(0) No(0)
• Log base a p can be written as logp/loga (both having base 10)
  Hence the ans is 1
• 9 years agoHelpfull: Yes(0) No(2)
• [(loga+logp)(loga-logp)]/[(logp+loga)(logp-loga)]
  taking minus common
  (-)[(loga+logp)(loga-logp)]/[(logp+loga)(loga-logp)]
  we get -1
  so ans will be
  -1
  
• 8 years agoHelpfull: Yes(0) No(3)