Physics and Mathematics
Properties of Logarithmic Function
1. Concept Overview
Logarithmic properties are rules that help us:
- simplify logarithmic expressions
- combine or split logarithms
- solve logarithmic equations easily
These properties are extremely important for:
- algebraic manipulation
- calculus (limits and differentiation)
- competitive exams
2. Fundamental Logarithmic Properties
Let [a>0] and [a≠1], and let [x>0], [y>0].
Property 1: Product Rule
[\log_a (xy) = \log_a x + \log_a y]
Meaning:
The logarithm of a product equals the sum of logarithms.
Example:
[\log_2 (8×4) = \log_2 8 + \log_2 4]
Property 2: Quotient Rule
[\log_a \dfrac{x}{y} = \log_a x − \log_a y]
Meaning:
The logarithm of a division equals the difference of logarithms.
Example:
[\log_2 \dfrac{8}{5} = \log_2 8 − \log_2 5]
Property 3: Power Rule
[\log_a (x^n) = n\log_a x]
Meaning:
The power of a number comes outside as a multiplier.
Example:
[\log_2 (8^5) = 5\log_2 8]
Property 4: Logarithm of 1
[\log_a 1 = 0]
Reason:
Because [a^0 = 1].
Example:
[\log_2 1 = 0]
Property 5: Logarithm of the Base
[\log_a a = 1]
Reason:
Because [a^1 = a].
Example:
[\log_2 2 = 1]
Property 6: Change of Base Formula
[\log_a x = \dfrac{\log_b x}{\log_b a}]
Use:
Helps convert logarithms to a common base (usually base 10 or [e]).
Example:
[\log_2 5 = \dfrac{\log_b 5}{\log_b 2}]
3. Important Observations
- Logarithmic properties apply only when arguments are positive
- Properties do not apply to sums or differences inside logarithms
- Order of application is very important
4. Conceptual Questions with Solutions
1. Why is log of a product equal to sum of logs?
Because logarithm converts multiplication into addition, making calculations easier.
2. Why is log of a quotient equal to difference of logs?
Because division is the inverse of multiplication.
3. Why does power come outside the logarithm?
Because logarithm measures the exponent to which the base is raised.
4. Can logarithmic properties be applied to negative numbers?
No. Logarithms are defined only for positive arguments.
5. Why is [\log_a 1 = 0]?
Because any base raised to power zero gives 1.
6. Why is [\log_a a = 1]?
Because base raised to power one remains unchanged.
7. Do logarithmic properties change the value?
No. They only change the form, not the value.
8. Why is change of base formula required?
Because calculators support only limited bases.
9. Can we apply log rules to sums?
No. Logarithm of sum has no simple rule.
10. Are logarithmic rules valid for all bases?
Yes, provided the base is positive and not equal to 1.
11. Why must arguments be positive?
Because no real exponent of a positive base gives a negative number.
12. Can logarithmic rules simplify equations?
Yes. They help convert complex expressions into simpler ones.
13. Do these properties apply to all logarithmic functions?
Yes, within their domain restrictions.
14. Why is logarithm important in calculus?
Because it simplifies growth rates and limits.
15. Are logarithmic properties exam-important?
Yes. They are frequently tested in board and competitive exams.
5. FAQ / Common Misconceptions
1. [\log(x+y) = \log x + \log y]
False. No such property exists.
2. Logarithmic rules work for zero.
False. Logarithm of zero is undefined.
3. Change of base formula changes value.
False. It only changes the base.
4. Logarithmic rules are optional.
False. They are essential tools.
5. Power rule applies to sums.
False. It applies only to products or powers.
6. Logs can take negative arguments.
False. Arguments must be positive.
7. Base can be negative.
False. Base must be positive and not equal to 1.
8. All logs are base 10.
False. Base can be any valid positive number.
9. Properties work without conditions.
False. Domain conditions must be checked.
10. Logarithms are outdated.
False. They are fundamental in modern mathematics and science.
6. Practice Questions with Step-by-Step Solutions
 Question 1. Simplify [\log_2 (8×4)].
Step-by-Step Solution:
Use product rule: [\log_a (xy)=\log_a x+\log_a y].
[\log_2 8 + \log_2 4].
[\log_2 8 = 3] and [\log_2 4 = 2].
Conclusion:
[\log_2 (8×4) = 5]
 Question 2. Simplify [\log_5 \dfrac{25}{5}].
Step-by-Step Solution:
Use quotient rule: [\log_a \dfrac{x}{y}=\log_a x−\log_a y].
[\log_5 25 − \log_5 5].
[\log_5 25 = 2] and [\log_5 5 = 1].
Conclusion:
[\log_5 \dfrac{25}{5} = 1]
 Question 3. Simplify [\log_3 (9^2)].
Step-by-Step Solution:
Use power rule: [\log_a (x^n)=n\log_a x].
[2\log_3 9].
[\log_3 9 = 2].
Conclusion:
[\log_3 (9^2) = 4]
 Question 4. Evaluate [\log_7 1].
Step-by-Step Solution:
Use the property [\log_a 1 = 0].
Conclusion:
[\log_7 1 = 0]
 Question 5. Evaluate [\log_4 4].
Step-by-Step Solution:
Use the property [\log_a a = 1].
Conclusion:
[\log_4 4 = 1]
 Question 6. Simplify [\log_2 (16^3)].
Step-by-Step Solution:
Apply power rule: [3\log_2 16].
[\log_2 16 = 4].
Conclusion:
[\log_2 (16^3) = 12]
 Question 7. Simplify [\log_3 81 − \log_3 3].
Step-by-Step Solution:
Use quotient rule in reverse.
[\log_3 \dfrac{81}{3}].
[\dfrac{81}{3} = 27].
[\log_3 27 = 3].
Conclusion:
[\log_3 81 − \log_3 3 = 3]
 Question 8. Simplify [2\log_5 25].
Step-by-Step Solution:
Use reverse power rule: [n\log_a x = \log_a (x^n)].
[\log_5 (25^2)].
[25^2 = 625].
[\log_5 625 = 4].
Conclusion:
[2\log_5 25 = 4]
 Question 9. Express [\log_2 8 + \log_2 4] as a single logarithm.
Step-by-Step Solution:
Use product rule.
[\log_2 (8×4)].
[8×4 = 32].
[\log_2 32 = 5].
Conclusion:
[\log_2 8 + \log_2 4 = 5]
 Question 10. Simplify [\log_{10} 1000].
Step-by-Step Solution:
Write [1000 = 10^3].
[\log_{10} 10^3 = 3].
Conclusion:
[\log_{10} 1000 = 3]
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