2 can be expressed as a power of 3, 3 can be expressed as a power of 5. Do not live in the world that has only rational numbers; it is irrational to do that.
2 can be expressed as a power of 3, 3 can be expressed as a power of 5. Do not live in the world that has only rational numbers; it is irrational to do that.
Unlock a deep understanding of mathematics with our Module and Summary! Clear definitions, comprehensive discussions, relevant example problems, and step-by-step solutions will guide you through mathematical concepts effortlessly. Learn with a systematic approach and discover the magic in every step of your learning journey. Mathematics doesn't have to be complicated—let's make it simple and enjoyable!
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
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Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Model Attribute Check Company Auto PropertyCeline George
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![Logaritmos
1. Hallar el valor de “x” :
a. Log xX x
+ Log xXx
= 4
b. Log X Log x
= 4
c. Log (x + 3) = Log x + Log 3
d. 2 Log X = Log 16
e. 3 Log X + 2 Log 2 0 = Log 32
f. Log3 (x + 3) = 1
g. Log 4X3
= 3/2
h. Log (x – 1) (4x – 4) = 2
i. Log (3x + 1)9 = 2
j. Log x(2x)2
+ Log x(5x/4) = Logx5
k. Log 4x = 3/2
l. Log 35 – Log 36 = Log 3X
m. Log ½X = 2
Log 3 + Log 1/21 + Log 1/21
n.
2Log6Log3LogXLog
2622 −+=
o. Log 7 (x – 1) – Log 7 (x – 5) = 1
p. Log X = Log 354 + Log 69 – Log 1357
q. Log X = ½ log 16 – 1/3 Log 8 + 1
r.
2,1log0114XLog7XLog ++++
2. Simplificar:
P = ½ [Log (x + y) + Log (x - y)] – Log Y
3. Hallar “A”, sabiendo que: m.n3
= p – 1
A = ½ Log m + 3/2 Log n + ½ p
4. Calcular:
E = Log (75/16) – 2 Log (5/9) – Log
(32/243)
5. Reducir: )zy/x(Log 523
6. Calcular M, en:
Log M = Log n + 2 Log p – Log q
7. Calcular “A”, en:
2 Log (A/2) = 3 Log A – Log 32
8. Hallar el valor de “X”:
16Log216Log5Log3XLog 4253 +=+
9. Hallar el valor de X :
)x2(Log)3x2(Log)7x(Log
1052 52 =− ++
10. Si log 3 = 0,4771213 ; Log 4 = 0,60203 ;
Hallar Log 48.
11. Si Log x2
= a y Log Y2
= b ;
Hallar 20 Log 10
y/x
12. Efectuar P = 8
Log (Log 2 2)
Logaritmos
1. Hallar el valor de “x” :
a. Log xX x
+ Log xXx
= 4
b. Log X Log x
= 4
c. Log (x + 3) = Log x + Log 3
d. 2 Log X = Log 16
e. 3 Log X + 2 Log 2 0 = Log 32
f. Log3 (x + 3) = 1
g. Log 4X3
= 3/2
h. Log (x – 1) (4x – 4) = 2
i. Log (3x + 1)9 = 2
j. Log x(2x)2
+ Log x(5x/4) = Logx5
k. Log 4x = 3/2
l. Log 35 – Log 36 = Log 3X
m. Log ½X = 2
Log 3 + Log 1/21 + Log 1/21
n.
2Log6Log3LogXLog
2622 −+=
o. Log 7 (x – 1) – Log 7 (x – 5) = 1
p. Log X = Log 354 + Log 69 – Log 1357
q. Log X = ½ log 16 – 1/3 Log 8 + 1
r.
2,1log0114XLog7XLog ++++
2. Simplificar:
P = ½ [Log (x + y) + Log (x - y)] – Log Y
3. Hallar “A”, sabiendo que: m.n3
= p – 1
A = ½ Log m + 3/2 Log n + ½ p
4. Calcular:
E = Log (75/16) – 2 Log (5/9) – Log
(32/243)
5. Reducir: )zy/x(Log 523
6. Calcular M, en:
Log M = Log n + 2 Log p – Log q
7. Calcular “A”, en:
2 Log (A/2) = 3 Log A – Log 32
8. Hallar el valor de “X”:
16Log216Log5Log3XLog 4253 +=+
9. Hallar el valor de X :
)x2(Log)3x2(Log)7x(Log
1052 52 =− ++
10. Si log 3 = 0,4771213 ; Log 4 = 0,60203 ;
Hallar Log 48.
11. Si Log x2
= a y Log Y2
= b ;
Hallar 20 Log 10
y/x
12. Efectuar P = 8
Log (Log 2 2)
Logaritmos
1. Hallar el valor de “x” :
a. Log xX x
+ Log xXx
= 4
b. Log X Log x
= 4
c. Log (x + 3) = Log x + Log 3
d. 2 Log X = Log 16
e. 3 Log X + 2 Log 2 0 = Log 32
f. Log3 (x + 3) = 1
g. Log 4X3
= 3/2
h. Log (x – 1) (4x – 4) = 2
i. Log (3x + 1)9 = 2
j. Log x(2x)2
+ Log x(5x/4) = Logx5
k. Log 4x = 3/2
l. Log 35 – Log 36 = Log 3X
m. Log ½X = 2
Log 3 + Log 1/21 + Log 1/21
n.
2Log6Log3LogXLog
2622 −+=
o. Log 7 (x – 1) – Log 7 (x – 5) = 1
p. Log X = Log 354 + Log 69 – Log 1357
q. Log X = ½ log 16 – 1/3 Log 8 + 1
r.
2,1log0114XLog7XLog ++++
2. Simplificar:
P = ½ [Log (x + y) + Log (x - y)] – Log Y
3. Hallar “A”, sabiendo que: m.n3
= p – 1
A = ½ Log m + 3/2 Log n + ½ p
4. Calcular:
E = Log (75/16) – 2 Log (5/9) – Log
(32/243)
5. Reducir: )zy/x(Log 523
6. Calcular M, en:
Log M = Log n + 2 Log p – Log q
7. Calcular “A”, en:
2 Log (A/2) = 3 Log A – Log 32
8. Hallar el valor de “X”:
16Log216Log5Log3XLog 4253 +=+
9. Hallar el valor de X :
)x2(Log)3x2(Log)7x(Log
1052 52 =− ++
10. Si log 3 = 0,4771213 ; Log 4 = 0,60203 ;
Hallar Log 48.
11. Si Log x2
= a y Log Y2
= b ;
Hallar 20 Log 10
y/x
12. Efectuar P = 8
Log (Log 2 2)
Logaritmos
1. Hallar el valor de “x” :
a. Log xX x
+ Log xXx
= 4
b. Log X Log x
= 4
c. Log (x + 3) = Log x + Log 3
d. 2 Log X = Log 16
e. 3 Log X + 2 Log 2 0 = Log 32
f. Log3 (x + 3) = 1
g. Log 4X3
= 3/2
h. Log (x – 1) (4x – 4) = 2
i. Log (3x + 1)9 = 2
j. Log x(2x)2
+ Log x(5x/4) = Logx5
k. Log 4x = 3/2
l. Log 35 – Log 36 = Log 3X
m. Log ½X = 2
Log 3 + Log 1/21 + Log 1/21
n.
2Log6Log3LogXLog
2622 −+=
o. Log 7 (x – 1) – Log 7 (x – 5) = 1
p. Log X = Log 354 + Log 69 – Log 1357
q. Log X = ½ log 16 – 1/3 Log 8 + 1
r.
2,1log0114XLog7XLog ++++
2. Simplificar:
P = ½ [Log (x + y) + Log (x - y)] – Log Y
3. Hallar “A”, sabiendo que: m.n3
= p – 1
A = ½ Log m + 3/2 Log n + ½ p
4. Calcular:
E = Log (75/16) – 2 Log (5/9) – Log
(32/243)](https://image.slidesharecdn.com/logaritmos-5to-181111044601/85/Logaritmos-5to-1-320.jpg)
