1 / 13

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:.

vladimir-osborn
Télécharger la présentation

log 3

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related