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Clase 117

Clase 117. Ecuaciones logarítmicas. Revisión del estudio individual. Ejercicio 6 ( e, q , h, k) pág. 13 L.T. Onceno grado. Comprobación:. e) log 5 (6x – 1 ) = 1. 1= 1. M.I: log 5 (6·1– 1). 6x – 1 = 5 1. x = 1. log 5 5 = 1.

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Clase 117

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  1. Clase 117 Ecuaciones logarítmicas

  2. Revisión del estudio individual Ejercicio 6 (e, q, h, k) pág. 13 L.T. Onceno grado Comprobación: e) log5(6x – 1 ) = 1 1= 1 M.I: log5(6·1– 1) 6x – 1= 51 x = 1 log5 5 = 1 Si logab = logac entonces b=c q) log3( –x2 + 5x) = log36 (b > 0, c > 0 a > 0, a  1) –x2 + 5x = 6 x2 – 5x + 6 = 0

  3. x2 – 5x + 6 = 0 x1= 3; x2= 2 (x – 3)(x – 2) = 0 Comprobación: Parax1= 3 M.I:log3( –x2 + 5x) = log3 –(3)2+5(3) = log3 (– 9+15) = log3 6 M.D: log3 6 Parax2 = 2 M.I:log3( –x2 + 5x) = log3 –(2)2+5(2) = log3 6 = log3 (– 4+10)

  4. logab = x si y solo si ax= b logab (a > 0, a  1, b > 0) x a = b Identidad fundamental logarítmica Si logab = logac entonces b=c (b > 0, c > 0 a > 0, a  1) loga1= 0 logaa = 1

  5. 3 2 log (x –x –x+2)= x b) 2 log x x 2 log (6x –7x+5) = 3 2 Ejercicio 1 Halla el conjunto solución de las siguientes ecuaciones logarítmicas: a) log52 c) log6– x x = 5

  6. 1 x1 = 3 2 log (6x –7x+5) = 3 3 x2 = 2 2 a) 6x2 –7x + 5 = 23 6x2 –7x + 5–8= 0 6x2 –7x –3= 0 (3x+1)(2x – 3)= 0 3 1 2 – 3 – 9 + 2= –7

  7. 2 1 1 – – log26 –7 + 5 3 3 1 7 = log26 + 5 + 9 3 7 2 = log2 +5 + 2 3 3 log (6x –7x+5) 2 Comprobación: 1 x1 = M.I: 3 M.D: 3 = log28 = 3

  8. 2 3 3 log26· –7 +5 2 2 3 2 1 3 2 S={ ;} log (6x –7x+5) 3 x2 = 2 2 Comprobación: M:I: = log2 ( 6·2,25–10,5 + 5) = log2 ( 13,5–10,5 + 5) = log2 8 = 3 M.D: 3

  9. 3 2 log (x –x –x+2)= x b) 2 log x x x3 – x2 – x + 2 = x2 x3 – 2x2 – x + 2 = 0 (x – 1)(x+1)(x – 2) = 0 Posibles raíces 1 – 2 – 1 2 x1= 1 x>0 N.S. indefine los logaritmos 1 1 – 1 – 2 x2= – 1 x1 1 – 1 – 2 0 x3 = 2

  10. 3 2 log (x –x –x+2)= x log (23– 22 –2 + 2) b) 2 2 log x x Comprobación: Para x = 2 3 2 M.I: log (x – x – x + 2) x = log24= 2 logxx2 log222 = log2 4= 2 M.D: S= {2} M.I = M.D

  11. loga b a = b log52 c) log6– x x = 5 log6– x x = 2 x = (6 – x )2 x = 36 – 12x + x2 0 = 36 – 13x + x2 (x – 9)(x –4) = 0 Indefine el logaritmo x1 = 9 ó x2 = 4 S= {4} Comp: x2 = 4 M.I: log6 – 4 4 = log24 = 2 M.D: 2

  12. log3( 2x –  x – 7) = log33  2x –  x – 7 = 3  2x – 3 =  x – 7 2x – 6 2x + 9 = x – 7 x + 16 = 6 2x Ejercicio 2: Para qué valores de x , se cumple que: 2 2

  13. 2 x + 16 = 6 2x 2 x2+ 32x+ 256 = 36(2x) x2+ 32x+ 256 = 72x x2 – 40x+ 256 = 0 (x – 32)(x – 8) = 0 x1= 32 ó x2= 8 Posibles soluciones ¡compruébalas!

  14. 1 – x2 f(x) = x2 Para el estudio individual 1. Sean las funciones: ; g(x) = sen x a) Halle de la forma más simple posible fog(x).. b) ¿Para qué valores de x está definida esta compuesta? 2. Ejercicio 7, incisos (a,b,k,l) página 14 del L.T de 11nogrado.

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