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All questions of Logarithms for SSC CGL Exam
The simplified form of log(75/16) -2 log(5/9) +log(32/343) is ?- a)log 2
- b)2 log 2
- c)log 3
- d)log 5
Correct answer is option 'A'. Can you explain this answer?
The simplified form of log(75/16) -2 log(5/9) +log(32/343) is ?
a)
log 2
b)
2 log 2
c)
log 3
d)
log 5
 | Ishaan Roy answered |
Steps to simplify log(75/16) - 2 log(5/9) + log(32/343):
1. Use logarithmic properties:
- log(a/b) = log(a) - log(b)
- log(a^b) = b*log(a)
2. Apply the properties:
- log(75/16) - 2 log(5/9) + log(32/343)
= log(75) - log(16) - 2(log(5) - log(9)) + log(32) - log(343)
3. Further simplify:
= log(75) - log(16) - 2log(5) + 2log(9) + log(32) - log(343)
= log(75) - log(16) - log(5^2) + log(9^2) + log(32) - log(343)
= log(75) - log(16) - log(25) + log(81) + log(32) - log(343)
= log(75*81*32) - log(16*25*343)
= log(194400) - log(137600)
= log(194400/137600)
= log(1.4142)
= log(2)
Therefore, the simplified form of log(75/16) - 2 log(5/9) + log(32/343) is log 2.
1. Use logarithmic properties:
- log(a/b) = log(a) - log(b)
- log(a^b) = b*log(a)
2. Apply the properties:
- log(75/16) - 2 log(5/9) + log(32/343)
= log(75) - log(16) - 2(log(5) - log(9)) + log(32) - log(343)
3. Further simplify:
= log(75) - log(16) - 2log(5) + 2log(9) + log(32) - log(343)
= log(75) - log(16) - log(5^2) + log(9^2) + log(32) - log(343)
= log(75) - log(16) - log(25) + log(81) + log(32) - log(343)
= log(75*81*32) - log(16*25*343)
= log(194400) - log(137600)
= log(194400/137600)
= log(1.4142)
= log(2)
Therefore, the simplified form of log(75/16) - 2 log(5/9) + log(32/343) is log 2.
The value of log23 x log32 x log34 x log43 is ?- a)1
- b)2
- c)3
- d)4
Correct answer is option 'A'. Can you explain this answer?
The value of log23 x log32 x log34 x log43 is ?
a)
1
b)
2
c)
3
d)
4
 | EduRev SSC CGL answered |
Given Exp.= log23 x log 32 x log34 x log43
= (log3 / log2) x ( log2 / log3) x (log4 / log3) x (log3 / log4) = 1
= (log3 / log2) x ( log2 / log3) x (log4 / log3) x (log3 / log4) = 1
If log 2 = 0.3010, then the number of digits in 264 is ?- a)18
- b)19
- c)20
- d)21
Correct answer is option 'C'. Can you explain this answer?
If log 2 = 0.3010, then the number of digits in 264 is ?
a)
18
b)
19
c)
20
d)
21
 | Ssc Cgl answered |
Required answer = [64 log10 2] + 1
= [ 64 x 0.3010 ] + 1
= 19.264 + 1
= 19 + 1
= 20
= [ 64 x 0.3010 ] + 1
= 19.264 + 1
= 19 + 1
= 20
Given that log10 2 = 0.3010, then log2 10 is equal to ?- a)0.3010
- b)0.6990
- c)1000 / 301
- d)699 / 301
Correct answer is option 'C'. Can you explain this answer?
Given that log10 2 = 0.3010, then log2 10 is equal to ?
a)
0.3010
b)
0.6990
c)
1000 / 301
d)
699 / 301
| | Ishaan Roy answered |
Understanding the Problem
To find log2 10, we can use the change of base formula. The relationship between logarithms of different bases is given by:
- log_a b = log_c b / log_c a
In this case, we can express log2 10 using base 10:
- log2 10 = log10 10 / log10 2
Now, we know that log10 10 = 1.
Calculating log2 10
Substituting the values we have:
- log2 10 = 1 / log10 2
Given that log10 2 = 0.3010, we can plug this value in:
- log2 10 = 1 / 0.3010
Next, we need to compute this division:
- log2 10 = 1000 / 301
This gives us the exact value of log2 10.
Conclusion
The correct answer is indeed option 'C':
- log2 10 = 1000 / 301
This method illustrates how to convert logarithms from one base to another using the change of base formula effectively.
To find log2 10, we can use the change of base formula. The relationship between logarithms of different bases is given by:
- log_a b = log_c b / log_c a
In this case, we can express log2 10 using base 10:
- log2 10 = log10 10 / log10 2
Now, we know that log10 10 = 1.
Calculating log2 10
Substituting the values we have:
- log2 10 = 1 / log10 2
Given that log10 2 = 0.3010, we can plug this value in:
- log2 10 = 1 / 0.3010
Next, we need to compute this division:
- log2 10 = 1000 / 301
This gives us the exact value of log2 10.
Conclusion
The correct answer is indeed option 'C':
- log2 10 = 1000 / 301
This method illustrates how to convert logarithms from one base to another using the change of base formula effectively.
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