Treść zadania

Najlepsze rozwiązanie

• 

  1 0

  antekL1 20.2.2013 (15:18)

  Oznaczenia:
  log_p(X) to "logarytm o podstawie p z X" oraz: a^X to "a do potęgi X.
  
  a)
  log_4 [log_10(100)] = log_4(2) = 1/2 [gdyż 4^(1/2) = pierwiastek(4) = 2 ]
  
  b)
  log_8 [log_pierwiastek(3)(9)] + log_8 [log_pierwiastek(2)(2)] =
  = log_8(4) + log_8(2) = log_8(8) = 1
  (gdyż 9 = [pierwiastek(3)]^4 oraz 2 = [pierwiastek(2)]^2, a także wzór na sumę logarytmów)
  
  c)
  25^[log_5(2)] - 2^[2+2log_2(3)] = { 5^[log_5(2)] }^2 - 4 * { 2^[log_2(3)] }^2 =
  = 2^2 - 4*3^2 = -32
  
  ------------
  Wzór na zamianę podstaw logarytmu: log_a(X) = [log_b(X)] / [log_b(a)]
  ------------
  
  d)
  log_4(81) * log_3(64) = { [log_3(81)] / [log_3(4)] } * { [log_4(64)] / [log_4(3)] } =
  = { 4 / [log_3(4)] } * { 3 / [log_4(3)] } = 12 / { [log_4(4)] / [log_4(3)] * log_4(3) } = 12
  
  e)
  log_5(16) * log_4(27) * log_3(25) =
  = { [log_4(16)] / [log_4(5)] } * { [log_3(27)] / [log_3(4)] } * { [log_5(25)] / [log_5(3)] } =
  = 2 * 3 * 2 / { [log_4(5)] * [log_3(4)] * [log_5(3)] } =
  12 / { [log_5(5)] / [log_5(4)] * [log_5(4)] / [log_5(3)] * [log_5(3)] } = 12
  
  f)
  log_9(16) * log_pierwiastek(2)(27) =
  = { [log_2(16) / [log_2(9)] } * { [log_3(27)] / [log_3(pierwiastek(2))] } =
  = 4 * 3 / { [log_2(9)] * [log_3(pierwiastek(2))] =
  = 12 / { [log_3(9)] / [log_3(2)] * (1/2) * [log_3(2)] } = 12 / [2 * (1/2)] = 12
  [gdyż log(pierwiastek(X)) = (1/2) log(X) ]

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