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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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slide1

Clase 119

Propiedades de los logaritmos

log3

= x

81

x = 4

x

4

3

= 81

slide2

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

slide3

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

slide4

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)

slide5

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

slide6

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

slide7

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

slide8

[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

slide9

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

slide10

1

a =

b

a) log10a + log10b = 0

log10(a·b) = 0

Como loga1 = 0

entoncesa·b = 1

a y b tienen que ser recíprocos.

slide11

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

slide12

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

slide13

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.