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log 3

Clase 119. Propiedades de los logaritmos. log 3. = x. 81. x = 4. x. 4. 3. = 81. log a b = x si y solo si a x = b. log a b. (a > 0 , a  1 , b > 0 ). x. a. = b. Identidad fundamental logarítmica. log a 1 = 0. log a a = 1. 1. log 3 81. 4. d) log 81. 3 4. Ejercicio:.

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log 3

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  1. Clase 119 Propiedades de los logaritmos log3 = x 81 x = 4 x 4 3 = 81

  2. logab = x si y solo si ax= b logab (a > 0, a  1, b > 0) x a = b Identidad fundamental logarítmica loga1= 0 logaa = 1

  3. 1 log381 4 d)log81 34 Ejercicio: Calcula y compara los resultados de la columna A y B A B a)log2(4·8) b)2log28 log216 – log28 c)log2(16:8) log24 + log28 log282

  4. b b) loga= loga b – logac c 1 e) log b = logab x ax Propiedades de los logaritmos Si a>0, b>0, c>0 tal que a1 entonces, se cumple: a) loga(b·c) = logab + logac c) loga bx = x logab (c  1) d) loga c · logc b = loga b (x  0)

  5. loga b + loga c a a) loga(b·c) = logab + logac Demostración: Se tiene que : loga b loga c a = a · =b·c (por definición) loga(b·c) = logab + logac l.q.q.d

  6. b) loga 7 + 3 loga 2 – loga16 1 c) (loga 3,2 + loga 40 – loga2) 1 2 3 Ejercicio 1 Expresa como un solo logaritmo: a>0, a1 a) loga 50 + loga 6 – loga 15

  7. b) loga 7 + 3 loga 2 – loga16 7· 23 1 = loga 2  16 a) loga 50 + loga 6 – loga 15 50 · 6 300 = loga = loga 15 15 = loga 20 2 7· 8 = loga 4 = loga 14

  8. [loga (3,2 · 40 : 2)] = 3  3,2 ·40 =loga 2 3  64 = loga 1 1 3 3 c) (loga 3,2 + loga 40 – loga2) = loga4

  9. Ejercicio 2 ¿Cuál es la relación que existe entre a y b (a>0, a1, b >0) para que: a) log10a + log10b = 0 b) log10 b =log10a – log10 5 c) log10 a·loga b =log10a3

  10. 1 a = b a) log10a + log10b = 0 log10(a·b) = 0 Como loga1 = 0 entoncesa·b = 1 a y b tienen que ser recíprocos.

  11. b) log10 b = log10a – log10 5 log10 b = log10 (a:5) b = a:5 a = 5b luego b es la quinta parte de a ó aes el quíntuplo de b

  12. 3 a= b c) log10 a · loga b = log10a3 log10b = log10 a3 b = a3 luego bes el cubo dea ó aes la raíz cúbica de b

  13. Para el estudio individual 1. Ejercicio 12 (a – e) pág.51 L.T. Onceno grado. Sabiendo que log103 = 0,477 Calcula: log1030; log103000; log100,003 2.

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