What is the approximate value of 'e'?
physics -Basic Mathematics(basic-L1)

Quiz
•
Physics
•
11th Grade
•
Medium
Mohit Gurjar
Used 5+ times
FREE Resource
10 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
e = 1.987
e = 2.496
e = 2.7183
e = 3.336
Answer explanation
The approximate value of 'e' is 2.7183, so the correct choice is e = 2.7183.
2.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Which of the following is correct expression ?
log(ab) = loga + logb
log(ab) = loga - logb
log(ab) = loga.logb
log(ab) = loga/logb
Answer explanation
The correct expression is log(ab) = loga + logb, which follows the logarithmic property of addition when multiplying two numbers.
3.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Which of the following is the correct expression ?
log(ab) = loga+logb
log(ab) = loga.logb
log(a/b) = loga-logb
log(ab) = loga/logb
Answer explanation
The correct expression is log(ab) = loga-logb because when you multiply two numbers in logarithms, it becomes subtraction.
4.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Which of the following is the correct value of ln(e)?
π
1
0
2
Answer explanation
The correct value of ln(e) is 1 because the natural logarithm of e is always equal to 1.
5.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
Find the value of log6 ( if log3=0.4771 , log2=0.3010).
0.6020
0.3010
0.7781
0.4771
Answer explanation
To find log6, we use the change of base formula: log6 = log(6)/log(10). Given log3=0.4771 and log2=0.3010, we can calculate log6 as log6 = log(2*3)/log(10) = (log2 + log3)/log(10) = (0.3010 + 0.4771)/0.3010 = 0.7781.
6.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
What is the value of log300? (if log3 = 0.4771)
3.4771
2.4771
2.4770
0.4771
Answer explanation
The value of log300 can be calculated by using the property of logarithms: log(a*b) = log(a) + log(b). Therefore, log300 = log(3*100) = log3 + log100 = 0.4771 + 2 = 2.4771.
7.
MULTIPLE CHOICE QUESTION
3 mins • 1 pt
What is value of log10-5?
-5
0
1
10
Answer explanation
The value of log10^-5 is -5. In logarithms, a negative exponent results in the reciprocal of the base raised to the positive exponent. Therefore, log10^-5 = log(1/10^5) = -5.
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