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LOGARITMOS
EJERCICIOS BÁSICOS :
1.- log864 = 2
2.-Calcular: log381 + log749 = 4 + 2 = 6
3.- Calcular :
resolución
𝐿𝑜𝑔 3
81
𝐿𝑜𝑔 3
81
= 𝐿𝑜𝑔
3
1
2
34
=
4
1
2
𝐿𝑜𝑔3
3
11
= 8 1. = 8
4.- Calcular : 𝐿𝑜𝑔4
2
32
resolución
𝐿𝑜𝑔4
2
32
=
𝑙𝑜𝑔
2
1
4
25
=
5
1
1
4
𝐿𝑜𝑔2
2
11
= 20 . 1 = 20
5.- Calcular : 3log3
7
resolución
3log3
7
= 7
6.- Calcular :
27 𝐿𝑜𝑔3
5
resolución
27 𝐿𝑜𝑔3
5
=
33 𝑙𝑜𝑔3
5
= 3𝑙𝑜𝑔3
5 3
= 5 3
= 125
 
 log 3x 277 2 x 9

 7.- Calcular :
resolución
 
 log 3x 277 2 x 9

 
3x-2 = 2 x + 18
X = 20
𝒃𝐥𝐨𝐠 𝒃 𝑵= N
1)8(Log
3
1
)16(Log
2
1
)x(Log 8.- Calcular :
resolución
1)8(Log
3
1
)16(Log
2
1
)x(Log      1816log)( 3
1
2
1
 LogxLog
    124log)(  LogxLog
    )10log(24log)(  LogxLog
log 10 = 1
log 𝑏 𝐴 + log 𝑏 𝐵 = log 𝑏 𝐴. 𝐵
log 𝑏 𝐴 − log 𝑏 𝐵 = log 𝑏
𝐴
𝐵
log 𝑏 𝐴 − log 𝑏 𝐵 + log 𝑏 𝐶 = log 𝑏
𝐴. 𝐶
𝐵 






2
10.4
log)(xLog
x = 20
9.- Calcular : Hallar “x” en: Logx = Log25 . Log52
resolución
Logx = Log25 . Log52
Logba . Logcb . Logdc = Logda
Aplicamos la regla de la cadena:
Log x = log5 5
Log x = 1 X = 101
= 10
10.- Calcular : Si: Log2 = 0,3
Log3 = 0,4
Hallar el valor de: E = Log 9 + Log 4 + Log6
resolución E = Log 9 + Log 4 + Log6
E = Log32 + Log 22 + Log(3 . 2 )
E = 2.Log3 + 2 .Log 2 + Log3+ log 2
Reemplazando:
E = 2.0,4 + 2 .0,3 + 0,4 + 0,3 = 1,54

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Logaritmos

  • 1. LOGARITMOS EJERCICIOS BÁSICOS : 1.- log864 = 2 2.-Calcular: log381 + log749 = 4 + 2 = 6 3.- Calcular : resolución 𝐿𝑜𝑔 3 81 𝐿𝑜𝑔 3 81 = 𝐿𝑜𝑔 3 1 2 34 = 4 1 2 𝐿𝑜𝑔3 3 11 = 8 1. = 8 4.- Calcular : 𝐿𝑜𝑔4 2 32 resolución 𝐿𝑜𝑔4 2 32 = 𝑙𝑜𝑔 2 1 4 25 = 5 1 1 4 𝐿𝑜𝑔2 2 11 = 20 . 1 = 20
  • 2. 5.- Calcular : 3log3 7 resolución 3log3 7 = 7 6.- Calcular : 27 𝐿𝑜𝑔3 5 resolución 27 𝐿𝑜𝑔3 5 = 33 𝑙𝑜𝑔3 5 = 3𝑙𝑜𝑔3 5 3 = 5 3 = 125    log 3x 277 2 x 9   7.- Calcular : resolución    log 3x 277 2 x 9    3x-2 = 2 x + 18 X = 20 𝒃𝐥𝐨𝐠 𝒃 𝑵= N
  • 3. 1)8(Log 3 1 )16(Log 2 1 )x(Log 8.- Calcular : resolución 1)8(Log 3 1 )16(Log 2 1 )x(Log      1816log)( 3 1 2 1  LogxLog     124log)(  LogxLog     )10log(24log)(  LogxLog log 10 = 1 log 𝑏 𝐴 + log 𝑏 𝐵 = log 𝑏 𝐴. 𝐵 log 𝑏 𝐴 − log 𝑏 𝐵 = log 𝑏 𝐴 𝐵 log 𝑏 𝐴 − log 𝑏 𝐵 + log 𝑏 𝐶 = log 𝑏 𝐴. 𝐶 𝐵        2 10.4 log)(xLog x = 20
  • 4. 9.- Calcular : Hallar “x” en: Logx = Log25 . Log52 resolución Logx = Log25 . Log52 Logba . Logcb . Logdc = Logda Aplicamos la regla de la cadena: Log x = log5 5 Log x = 1 X = 101 = 10 10.- Calcular : Si: Log2 = 0,3 Log3 = 0,4 Hallar el valor de: E = Log 9 + Log 4 + Log6 resolución E = Log 9 + Log 4 + Log6 E = Log32 + Log 22 + Log(3 . 2 ) E = 2.Log3 + 2 .Log 2 + Log3+ log 2 Reemplazando: E = 2.0,4 + 2 .0,3 + 0,4 + 0,3 = 1,54