if z=(log(base a)(log(base b)(a)))/(log(base b)(log(base a)(b)))
simplify z

• Answer z= - log(base a)(b)
• 10 years agoHelpfull: Yes(12) No(0)
• Z = [ log{log(a)/log(b)}/log(a) ] / [log{log(b)/log(a)}/log(b)]
  Z= (log a)^3 / (log b)^3
  Z = 3 log a / 3 log b
  so ans is Z = log (base b) a
• 10 years agoHelpfull: Yes(12) No(3)
• Answer: z = log(base a) (1/b)
  
  
  Formula:
  1)Log(base a)(b) = log(b)/log(a)
  2)Log(x/y) = logx - logy
  
  We have,
  Z = [ log{log(a)/log(b)}/log(a) ] / [log{log(b)/log(a)}/log(b)]
  
  As log(logb / loga) = log(logb) - log(loga)
  
  => z = [logb / loga] * [ {log(loga) - log(logb)} / {log(logb) - log(loga)}]
  => z = -log(b) / log(a)
  => z = log(base a) (1/b)
• 10 years agoHelpfull: Yes(5) No(2)
• We cannot directly use the log a / log b form as the given problem is log of log format.
  
  One way of solving the problem is assuming the relation between a and b.
  
  Let a = b power n
  a = b ^ n
  
  Now the numerator is loga(logba) = loga(logb(b^n))
  
  As the power goes out it will become loga(nlogbb)
  As logbb=1, the entire numerator can be simplified as loga(n)
  
  Now the denominator is logb(logab)
  As we assumed a = b ^ n, denominator can be written as
  = logb(logb^n(b))
  
  As the power is for the base, it will become reciprocal and goes out it will become logb(1/nlogbb)
  As logbb=1, the entire numerator can be simplified as logb(1/n)
  As 1/n can be written as n power –1, the denominator can be simplified as –logb(n)
  
  Now, we can use loga / log b form for further simplification.
  
  Numerator is loga(n) = log n/ log a
  Denominator is –logb(n) = –log n/ log b
  
  Now by cancelling out log n in numerator and denominator, we get –log b/ log a
  
  This can be written as –loga(b)
  
  –loga(b) is the answer.
  
• 10 years agoHelpfull: Yes(4) No(0)
• log a^ (log b^a))/log b^ ( log a^b)
  (log b^a) (log b) log a log b * log a
  = --------------- * --------------- = ---------------- * -----------------------
  log a (log a^b) log b* log a log a
  
  =1
• 10 years agoHelpfull: Yes(0) No(1)
• z=(log(base a)(log(base b)(a)))/(log(base b)(log(base a)(b)))
  => z=[log(base b)(a))/log(a)]*[log(b)/(log(base a)(b))]
  => z=[log(a)/{log(a)*log(b)}]*[{log(b)*log(a)}/log(b)]
  => Z=log(a)/Log(b)
  => Z=log(base b) (a) .....Ans
• 9 years agoHelpfull: Yes(0) No(1)