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Clase 120

Clase 120. Ejercicios sobre propiedades de los logaritmos. Revisión del estudio individual. Sabiendo que log 10 3 = 0,477 Calcula: log 10 30; log 10 3000; log 10 0,003. log 10 30 = log 10 (3·10). = log 10 3 + log 10 10. = 0,477 + 1. = 1,477. log 10 3000 ; log 10 0,003.

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Clase 120

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  1. Clase 120 Ejercicios sobre propiedades de los logaritmos

  2. Revisión del estudio individual Sabiendo que log103 = 0,477 Calcula: log1030; log103000; log100,003 log1030 = log10 (3·10) = log10 3 + log10 10 = 0,477 + 1 = 1,477

  3. log103000 ; log100,003 log103000 = log10 (3 · 1000) = log10 3 + log10 1000 = log10 3 + log10 103 = 0,477 + 3 = 3,477 log10 (3 :1000) log 10 0,003 = = log10 3 –log10 103 = 0,477 – 3 = – 2,523

  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 logc b = loga c log2128 7 = 3 log2 8 d) loga c · logc b = loga b ( cambio de base) Ejemplo: log8 128 =

  6. 7 c) log2 0,064 Ejercicio 1 Sabiendo que log2 10 = 3,32 Calcula: a) log2 1,6 b) log2 0,008 Estudio Individual

  7. 16 a) log2 1,6 = log2 10 log2 10 = 3,32 = log2 16 – log210 = 4 – 3,32 = 0,68 b) log2 0,008 = log2 (8·10 -3 ) = log28 + log2 10 -3 = 3 + (– 3· 3,32) = –6,96 = 3 + (– 9,96)

  8. b) log5 N 4 25  N Ejercicio 2 : Si log5 N = k, expresa en función de k los siguientes logaritmos: a) log5125N c) log5 d) logN 5

  9. 1 1 = log5 N k b) log5 N 4 25 1 1  N 4 4 k = 4 log5 5 d) logN 5 log5 N = k = = log5 N = log5 125 + log5N a) log5125N = 3 + k = log5 N – log525 = k – 2 c) log5 = log5 N = log5 N

  10. 1 3 1 1 log3 b a) log3x = log3 b– log3 c + 4 2 4 2 3 log2 a log2 x = log2 b)  a2 – Ejercicio 3 Resuelve considerando todas las expresiones positivas:

  11. log3 x =log 3 b– log 3 3 1 1 log3 b a) log3x = log3 b– log3 c + 4 2 4 b  c  c  c  c  c  c b  c b x = = c c log3 x =log 3 b– log 3 c 0,5 b log 3 x = log3 .

  12. log2 a log2 x = log2 – b) 1  a  a 2 6 6 6 3 3 6 6 3  a3  a2  a2  a4  a  a  a  a2 log2 x =log2 log2 x =log2 = x = =

  13. 1 7 Para el estudio individual Resuelve las ecuaciones: log7 (2x2 – 5x) =1 a) x1 = 3,5 x2 = –1 b) log2(x – 1) 7 : 7 = x = 2

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