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3. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
𝐷𝑒𝑓𝑖𝑛𝑖𝑟 𝑦 𝑟𝑒𝑐𝑜𝑛𝑜𝑐𝑒𝑟
𝑙𝑎𝑠 𝑝𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑𝑒𝑠
𝑑𝑒𝑙 𝑙𝑜𝑔𝑎𝑟𝑖𝑡𝑚𝑜
𝑅𝑒𝑠𝑜𝑙𝑣𝑒𝑟 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠 𝑒
𝑖𝑛𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛𝑒𝑠
𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎𝑠
𝐼𝑑𝑒𝑛𝑡𝑖𝑓𝑖𝑐𝑎𝑟 𝑙𝑎 𝑔𝑟á𝑓𝑖𝑐𝑎 𝑑𝑒
𝑢𝑛𝑎 𝑓𝑢𝑛𝑐𝑖ó𝑛 𝑙𝑜𝑔𝑎𝑟í𝑡𝑚𝑖𝑐𝑎
𝑐𝑜𝑛 𝑠𝑢𝑠 𝑝𝑟𝑜𝑝𝑖𝑒𝑑𝑎𝑑𝑒𝑠
4. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Logaritmos
El concepto de logaritmo se remonta hasta los estudios
de Arquímedes y se origina a partir de la comparación
entre una progresión aritmética y una progresión
geométrica.
P.A. 1 2 3 4 5 6 7 8
P.G. 𝑎1
𝑎2
𝑎3 𝑎4
𝑎5 𝑎6 𝑎7
𝑎8
Los logaritmos tiene una gran importancia en la
ciencia y tiene diversas aplicaciones, por ejemplo:
La escala de decibeles el nivel de intensidad 𝐵
medido en decibeles se define:
𝐵 = 10log
𝐼
𝐼0
𝐼: intensidad sonora.
𝐼0 = 10−12 W/m2
5. C R E E M O S E N L A E X I G E N C I A
Logaritmos
1. Definición
log𝑏𝑥 = 𝑦
C U R S O D E Á L G E B R A
⟺ 𝑏𝑦
• log264 = 6
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
Sean los números 𝑥 > 0 y 𝑏 > 0 ∧ 𝑏 ≠ 1, entonces:
⟺ 26
= 64
• log381 = 4 ⟺ 34
= 81
• log5
1
25
= −2 ⟺ 5−2
=
1
25
• log88 = 1 ⟺ 81
= 8
log𝑏𝑏 = 1 log𝑏1 = 0
= 𝑥
donde:
𝑥 → número
𝑏 → base del logaritmo
𝑦 → logaritmo de 𝑥 en base 𝑏
• log31 = 0 ⟺ 30
= 1
𝐏𝐫𝐨𝐩𝐢𝐞𝐝𝐚𝐝𝐞𝐬:
2. Identidad fundamental
De la definición, se desprende que si 𝑥 > 0 y 𝑏 > 0 ∧
𝑏 ≠ 1, entonces:
𝒃
log𝒃𝑥
= 𝑥
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
• 𝟓log𝟓13 = 13 • 𝟕log𝟕𝜋 = 𝜋
; 𝑦 ∈ ℝ
6. C R E E M O S E N L A E X I G E N C I A
Teoremas sobre logaritmos
C U R S O D E Á L G E B R A
1. log𝑏𝑥. 𝑦 = log𝑏𝑥 + log𝑏𝑦
• log515 = log53 + log55
log5 3.5 =
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
• log37 + log34 = log328
log3 7.4 =
2. log𝑏
𝑥
𝑦
= log𝑏𝑥 − log𝑏𝑦
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
log2
5
3
= log25 − log23
= log5
7
2
log57 − log52
•
•
3. log𝑏𝑥 𝑛
= log𝑏𝑥
𝒏
• log55 7
= log55
𝟕
ቋ
1
= 7
• 𝟑.log2𝑥 = log2 𝑥𝟑
Sean 𝑥, 𝑦 ∈ ℝ+
y una base "𝑏" 𝑏 > 0 ∧ 𝑏 ≠ 1
• log832 =
• log81729 =
log2𝟑25
= . log22
𝟓
𝟑
log3𝟒36
= . log33
𝟔
𝟒
log𝑏𝑥 = log𝑏𝒏 𝑥𝑛
• log3
2
4 = log3
2
𝟑 43
= log226
𝐂𝐨𝐧𝐬𝐞𝐜𝐮𝐞𝐧𝐜𝐢𝐚𝐬
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
En general
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
= 6
log𝑏𝑥 = log𝑛
𝑏
𝒏
𝑥
=
5
3
=
3
2
• log12125
= log11 5
= log 121 25
log3
2
4
𝑏𝒎
log 𝑥 𝑛 . log𝑏𝑥
𝒏
𝒎
=
7. C R E E M O S E N L A E X I G E N C I A
log𝑏𝟓
log𝑏𝟑
log 𝒙
log 𝒃
𝒂
𝒂
𝐍𝐨𝐭𝐚𝐬:
log𝑏
𝑛
𝑥 = log𝑏𝑥 𝑛
log𝑏
𝑛
𝑥
log𝑏𝑥𝑛
≠
4.
log𝒃𝒙 =
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
𝐑𝐞𝐠𝐥𝐚 𝐝𝐞𝐥 𝐜𝐚𝐦𝐛𝐢𝐨 𝐝𝐞 𝐛𝐚𝐬𝐞
C U R S O D E Á L G E B R A
𝑖)
𝑖𝑖)
• log2
3
5= log25 3
• log3
2
81= log381 2
= 4 2 = 16
log2
3
𝑥
• log2𝑥3
≠
log5
2
25
• log5252
≠
𝟐
𝟐
• log𝟑𝟓 =
𝒂 > 0 ∧ 𝒂 ≠ 1
log𝟏𝟏𝟏𝟑
log𝟏𝟏𝟖
log𝟓16
log𝟓2
• = log216 = 4
• log𝟖𝟏𝟑 =
Consecuencias:
log𝒃𝒙 =
1
log𝒙𝒃
• log𝟖𝟑=
1
log𝟑𝟖
log𝟒𝟗
1
log𝟗𝟒
=
log𝑏𝑎. log𝑎𝑥 = log𝑏𝑥
• log27. log732 = log232 = 5
• • log35. log58. log881 = log381 = 4
𝐎𝐛𝐬𝐞𝐫𝐯𝐚𝐜𝐢ó𝐧
log𝑏𝑥𝑛
= 𝑛 log𝑏 𝑥
=
log𝟏𝟑
log𝟖
Si 𝑥 ∈ ℝ, 𝑏 > 0 ∧ 𝑏 ≠ 1
; 𝑛 es par
• log5 −5 4
= 4. log5 −5 = 4. log55 = 4
• log3𝑥2
= 2. log3 𝑥
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
8. C R E E M O S E N L A E X I G E N C I A
5. 𝐑𝐞𝐠𝐥𝐚 𝐝𝐞𝐥 𝐢𝐧𝐭𝐞𝐫𝐜𝐚𝐦𝐛𝐢𝐨
𝒙
log𝒃𝒚
= 𝒚
log𝒃𝒙
𝐄𝐣𝐞𝐦𝐩𝐥𝐨𝐬:
• 5log27
= 7log25
• 13log211
= 11log213
Aplicación:
Reducir la siguiente expresión:
𝑁 =
81log3 5 − 38log32
4log23 log432
Resolución :
𝑁 =
5log3 81 − 3log328
3log24 log432
𝑁 =
54
− 28
3log24 . log432
𝑁 =
369
3log232
𝑁 =
369
35
∴ 𝑁 =
41
27
C U R S O D E Á L G E B R A
9. C R E E M O S E N L A E X I G E N C I A
Sistemas de logaritmos
Sistema decimal (Briggs) Sistema natural (Neperiano)
Ejemplos: Ejemplos:
• logee
• Lne5
• log1010
• log1000
Es aquel sistema de logaritmos en
donde la base es 10.
log10𝑥 = log𝑥
Es aquel sistema de logaritmos en
donde la base es e = 2,7182 …
loge𝑥 = Ln𝑥
C U R S O D E Á L G E B R A
• 10log5
• eLn7
= 5 = 7
= log10
= 3
= Lne
= 5Lne
𝐍𝐨𝐭𝐚𝐜𝐢ó𝐧: 𝐍𝐨𝐭𝐚𝐜𝐢ó𝐧:
Existen infinitos sistemas
de logaritmos para una
base positiva y diferente
de 1.
Los más importantes son
el sistema decimal y el
natural.
= 1 = 1
= 5
10. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Ecuaciones logarítmicas
Son las ecuaciones donde la variable se encuentra
afectado del logaritmo
Ejemplos :
• log3 𝑥 − 5 = log3 2 − 𝑥
• log2 𝑥2
− 5 = 2
• log5 𝑥 = 7 − 𝑥
Para la resolución de las ecuaciones logarítmicas,
se debe tener en cuenta los pasos siguientes:
Paso 1
Garantizar la existencia de los logaritmos en
los reales
log𝑏 𝑥 ∈ ℝ
Paso 2
Despejar 𝑥: log𝑏 𝑥 = log𝑏 𝑦 ↔ 𝑥 = 𝑦
Paso 3 Intersectar los resultados anteriores
Si: 𝑥 > 0 y 𝑏 > 0 ∧ 𝑏 ≠ 1,
Propiedad :
log𝑏 𝑥 = log𝑏 𝑦 ↔ 𝑥 = 𝑦
Ejemplo: Resuelva la ecuación siguiente log𝑥 12 − 𝑥 = 2
Resolución :
• log𝑥 12 − 𝑥 existe en reales ↔ 12 − 𝑥 > 0
↔ 𝑥 ∈ 0; 12 − 1
• De la definición de logaritmos, se tiene:
𝑥2
= 12 − 𝑥 ↔ 𝑥2
+𝑥 − 12 = 0 ⟷ 𝑥 = −4 ∨ 𝑥 = 3
No cumple Si cumple
∴ 𝐶𝑆 = 3
↔ 𝑏 > 0 ∧ 𝑏 ≠ 1 ∧ 𝑥 > 0
∧ 𝑥 > 0 ∧ 𝑥 ≠ 1
11. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Gráfica de una función logarítmica.
Si x > 0 𝑦 𝑏 > 0 y 𝑏 ≠ 1, entonces la función exponencial
se define como
𝑓𝑥 = log𝑏𝑥
𝐄𝐣𝐞𝐦𝐩𝐥𝐨:
𝑓𝑥 = log2𝑥 ; 𝑥 ∈ ℝ+
Tabularemos algunos puntos para inducir la gráfica
𝑥 𝑦
8 3
4 2
2 1
1 0
−1
Τ
1 2
−2
Τ
1 4
𝑦
𝑥
𝐄𝐣𝐞𝐦𝐩𝐥𝐨:
𝑓𝑥 = log1
2
𝑥 ; 𝑥 ∈ ℝ+
Tabularemos algunos puntos para inducir la gráfica
𝑥 𝑦
8
4
2
2
1
1 0
−1
Τ
1 2
−2
Τ
1 4
−3
𝑦
𝑥
Dom𝑓=ℝ+
∧ Ran𝑓 = ℝ
1 2
1
4 8
2
3
1 2
−1
4
−2
8
−3
12. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
En general:
Sea 𝑓𝑥 = log𝑏 𝑥 ; 𝑏 ≠ 1 ∧ 𝑏 > 0 ∧ 𝑥 > 0
b> 1
𝑦
𝑥
𝑓𝑥 = log𝑏 𝑥
0 < 𝑏 < 1
𝑦
𝑥
1 1
𝑓𝑥 = log𝑏 𝑥
Ambas funciones son inyectivas
log𝑏 𝑥 = log𝑏 𝑦 ↔ 𝑥 = 𝑦
Función creciente
log𝑏 𝑥 < log𝑏 𝑦 ↔ 𝑥 < 𝑦
Función decreciente
log𝑏 𝑥 < log𝑏 𝑦 ↔ 𝑥 > 𝑦
No cambia el sentido de
la desigualdad
Si cambia el sentido de
la desigualdad
𝐄𝐣𝐞𝐦𝐩𝐥𝐨 ∶
Determine el dominio de la siguiente función
𝑓𝑥 = log𝑥(5 − 𝑥)
Resolución :
log𝑥 5 − 𝑥 existe en los ℝ
↔ 5 − 𝑥 > 0 ∧ 𝑥 > 0 ∧ 𝑥 ≠ 1
↔ 5 > 𝑥 ∧ 𝑥 > 0 ∧ 𝑥 ≠ 1
↔ 𝑥 ∈ 0; 5 − 1
∴ Dom𝑓 = 0; 5 − 1
13. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Inecuaciones logarítmicas
Son las inecuaciones donde la variable se encuentra
afectado del logaritmo
Ejemplos:
• log5 𝑥 − 2 ≤ log5 4 − 𝑥
• log2 2𝑥 − 1 < −1
• log3 𝑥 ≥ 5 − 𝑥
Resuelva la inecuación
log1
2
3𝑥 − 1 > −1
Resolución :
• log1
2
3𝑥 − 1 existe en los ℝ
↔ 3𝑥 − 1 > 0
log1
2
3𝑥 − 1 > −1 log1
2
1
2
3𝑥 − 1 < 2 𝑥 < 1
∩
∴ CS =
1
3
; 1
Para su resolución, se debe tener en cuenta los
pasos siguientes:
Paso 1
Garantizar la existencia de los
logaritmos en los reales
log𝑏 𝑥 ∈ ℝ ↔ 𝑏 > 0 ∧ 𝑏 ≠ 1 ∧ 𝑥 > 0
Paso 2
Despejar 𝑥:
Si 𝑏 > 1: log𝑏 𝑥 < log𝑏 𝑦 ↔ 𝑥 < 𝑦
Paso 3 Intersectar los resultados anteriores
Si 0 < 𝑏 < 1: log𝑏 𝑥 < log𝑏 𝑦 ↔ 𝑥 > 𝑦
Ejemplo:
⟷ 𝑥 >
1
3
• La base es menor a uno, el sentido de la
desigualdad si cambia
1
3
⟷ log1
2
3x − 1 > log1
2
2
⟷
14. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
Demostraciones
Identidad fundamental
Si 𝑥 > 0 y 𝑏 > 0 ∧ 𝑏 ≠ 1, entonces:
𝒃
log𝒃𝑥
= 𝑥
De la definición tenemos:
log𝑏𝑥 = 𝑦 ⟺ 𝑏𝑦
= 𝑥 ; 𝑦 ∈ ℝ
Reemplazamos el valor de 𝑦:
𝑏𝑦
= 𝑥
⟺ 𝒃
log𝒃𝑥
= 𝑥
1. log𝑏𝑥. 𝑦 = log𝑏𝑥 + log𝑏𝑦
De la definición tenemos:
log𝑏𝑥 = 𝑚 ⟺ 𝑥 = 𝑏𝑚
log𝑏𝑦 = 𝑛 ⟺ 𝑦 = 𝑏𝑛
Multiplicando:
𝑥. 𝑦 = 𝑏𝑚
. 𝑏𝑛
= 𝑏𝑚+𝑛
⟺ 𝑏𝑚+𝑛
= 𝑥. 𝑦
⟺ 𝑚 + 𝑛 = log𝑏 (𝑥. 𝑦)
൝
(∗)
De ∗ , nos queda
log𝑏𝑥 + log𝑏𝑦 = log𝑏(𝑥. 𝑦)
2. log𝑏
𝑥
𝑦
= log𝑏𝑥 − log𝑏𝑦
De la definición tenemos:
log𝑏𝑥 = 𝑚 ⟺ 𝑥 = 𝑏𝑚
log𝑏𝑦 = 𝑛 ⟺ 𝑦 = 𝑏𝑛
Dividiendo:
𝑥
𝑦
=
𝑏𝑚
𝑏𝑛
= 𝑏𝑚−𝑛
⟺ 𝑏𝑚−𝑛
=
𝑥
𝑦
⟺ 𝑚 − 𝑛 = log𝑏 (
𝑥
𝑦
)
൝
(∗)
De ∗ , nos queda
log𝑏𝑥 − log𝑏𝑦 = log𝑏(
𝑥
𝑦
)
15. C R E E M O S E N L A E X I G E N C I A
C U R S O D E Á L G E B R A
3. log𝑏𝑥 𝑛
= log𝑏𝑥
𝒏
De la definición tenemos:
log𝑏𝑥 = 𝑚 ⟺ 𝑥 = 𝑏𝑚
⟺ 𝑥 = 𝑏𝑚
𝑛 𝑛
⟺ 𝑥𝑛
= 𝑏𝑚.𝑛
⟺ 𝑚𝑛 = log𝑏 𝑥 𝑛
Como: 𝑚 = log𝑏 𝑥
⟺ log𝑏𝑥
𝑛 = log𝑏 𝑥 𝑛
log 𝒙
log 𝒃
𝒂
𝒂
4.
log𝒃𝒙 =
𝐑𝐞𝐠𝐥𝐚 𝐝𝐞𝐥 𝐜𝐚𝐦𝐛𝐢𝐨 𝐝𝐞 𝐛𝐚𝐬𝐞
𝒂 > 0 ∧ 𝒂 ≠ 1
De la definición tenemos:
log𝑏𝑥 = 𝑚 ⟺ 𝑥 = 𝑏𝑚
log𝑎𝑏 = 𝑛 ⟺ 𝑏 = 𝑎𝑛
Entonces:
𝑥 = 𝑏𝑚
= 𝑎𝑛 𝑚
= 𝑎𝑚.𝑛
⟺ 𝑥 = 𝑎𝑚.𝑛
⟺ 𝑚𝑛 = log𝑎 𝑥
൝
(∗)
De ∗ , nos queda
⟺ log𝑏𝑥. log𝑎𝑏 = log𝑎𝑥
Entonces:
log 𝒙
log 𝒃
𝒂
𝒂
log𝒃𝒙 =
𝐂𝐨𝐧𝐬𝐞𝐜𝐮𝐞𝐧𝐜𝐢𝐚:
Si: 𝒂 = 𝒙 tenemos
log𝒃𝒙 =
log 𝒙
log 𝒃
𝒙
𝒙 =
log 𝒃
𝒙
𝟏
16. w w w . a c a d e m i a c e s a r v a l l e j o . e d u . p e