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Clase 137

Clase 137. Función exponencial y logarítmica. Ejercicios. t + 2. ·. f (t) = 3. 1. 9 t. a) Halla el valor de t, tal que f(t) =  27. Ejercicio 1. Sean las funciones :. g(x) = log 6 x + log 6 (x – 1). b) Determina el dominio de f y g. c) Esboce el gráfico de f. t + 2. ·.

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Clase 137

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  1. Clase 137 Función exponencial y logarítmica. Ejercicios.

  2. t + 2 · f (t) = 3 1 9t a) Halla el valor de t, tal que f(t) =  27. Ejercicio 1 Sean las funciones : g(x) = log6x + log6(x – 1) b) Determina el dominio de f yg. c) Esboce el gráfico de f.

  3. t + 2 · f (t) = 3 1 1 = 27 9t 9t 3t +2 – 2t = 3 1 3 3 · 2 2 32t t + 2 · 3 3t +2 = 3 1,5 32 – t =3 t = 0,5 2 – t = 1,5

  4. t + 2 · f (t) = 3 1 9t b) Determina el dominio de f y g g(x) = log6x + log6(x – 1) Dom f: t x  0 y x 1 Dom g: x; x 1 0 1

  5. t – 2 f(t) = 1 3 1 f(t) = 3t+2 · 9t f(t) = 32 – t c) y f(t) = 3t+2 · 3–2t f(t) = 3t+2– 2t f(t) = 32– t 1 0 t 2

  6. Ejercicio 2 Sean las funciones: m(x) =log4(x + 5)2 h(x) = log2(x – 2 ) + 1 a) Calcula los ceros de h y esboza su gráfico. b) Determina los valores reales para los cuales m y h toman el mismo valor.

  7. h(x) = log2(x – 2 ) + 1 log2(x – 2 ) + 1 = 0 log2(x – 2 ) = – 1 x – 2 = 2– 1 x = 0,5 + 2 x = 2,5 El cero de la función h es x = 2,5.

  8. h(x) = log2(x – 2 ) + 1 y 2 1 x 0 2 3 1

  9. x + 5 log2 = 1 x – 2 1 1 2 2 log2(x – 2 ) + 1 log4(x + 5)2 = b) log2(x – 2 ) + 1 log2(x + 5)2 = 2 log2(x – 2 ) + 1 log2(x + 5)2 = log2(x – 2 )+ 1 log2 (x + 5)2 = log2(x + 5) – log2(x – 2) = 1

  10. log2 = 1 x + 5 x + 5 x – 2 x – 2 = 2 x + 5 = 2(x – 2) x + 5 = 2x – 4 x = 9 ¡Compruébala!

  11. Ejercicio 3 Dadas las funciones: p(x) = 2log(x – 2) · 40,5log(x– 2) k(x) = log3(x2 – 1) Determina los valores de x para los cuales p(x) = k(2).

  12. k(x) = log3(x2 – 1) k(2) = log3(22 – 1) k(2) = 1 k(2) = log33 p(x) = 2log(x – 2) · 40,5log(x– 2) 2log(x – 2) · 40,5log(x– 2) = 1 2log(x – 2) ·(2)2(0,5log(x– 2)) = 20 2log(x – 2)+ log(x– 2) = 20 22log(x – 2)= 20

  13. 22log(x – 2) = 20 2 log(x – 2) = 0 log(x – 2)= 0 x – 2 = 100 x – 2 = 1 x = 3

  14. 3.Dada la función : p(x) = log x2 + 2x Para el estudio individual 1.Ejercicio 19 pág. 55 L.T. Onceno grado. 2.Ejercicio 24 pág. 55 L.T. Onceno grado. Demuestra que (qop)(x)= x2 + 2x si q(x) = 102x.

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