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Quantitative Aptitude ยท Logarithms and Progressions

Logarithms - Formulas, Key Points and Examples

Explanation

๐Ÿ”ต LOGARITHMS ๐ŸŸข 1. BASIC CONCEPT OF LOGARITHM A logarithm is another way of expressing an exponential relationship. If:
ax=N a^x=N
then:
logโกaN=x \log_a N=x
Here: a = base N = argument x = logarithm ๐ŸŸข 2. CONDITIONS FOR A LOGARITHM For:
logโกaN \log_a N
the conditions are:
a>0 a>0
aโ‰ 1 a\neq1
and:
N>0 N>0
๐ŸŸข 3. BASIC LOGARITHM VALUES
logโกa1=0 \log_a1=0
because:
a0=1 a^0=1
Also:
logโกaa=1 \log_a a=1
because:
a1=a a^1=a
๐ŸŸข 4. LOGARITHM OF A POWER
logโกa(an)=n \log_a(a^n)=n
For example:
logโก2(25)=5 \log_2(2^5)=5
๐ŸŸข 5. PRODUCT RULE The logarithm of a product is the sum of the logarithms.
logโกa(xy)=logโกax+logโกay \log_a(xy) = \log_a x+\log_a y
๐ŸŸข 6. QUOTIENT RULE The logarithm of a quotient is the difference of the logarithms.
logโกa(xy)=logโกaxโˆ’logโกay \log_a\left(\frac{x}{y}\right) = \log_a x-\log_a y
๐ŸŸข 7. POWER RULE The power can be brought in front of the logarithm.
logโกa(xn)=nlogโกax \log_a(x^n) = n\log_a x
๐ŸŸข 8. ROOT RULE Since:
xn=x1/n \sqrt[n]{x}=x^{1/n}
we get:
logโกaxn=1nlogโกax \log_a\sqrt[n]{x} = \frac{1}{n}\log_a x
๐ŸŸข 9. CHANGE OF BASE FORMULA The logarithm can be changed from one base to another.
logโกax=logโกbxlogโกba \log_a x = \frac{\log_b x}{\log_b a}
A commonly used form is:
logโกax=lnโกxlnโกa \log_a x = \frac{\ln x}{\ln a}
๐ŸŸข 10. COMMON LOGARITHM A logarithm with base 10 is called a common logarithm.
logโก10x \log_{10}x
Usually the base 10 is omitted:
logโกx \log x
๐ŸŸข 11. NATURAL LOGARITHM A logarithm with base e is called a natural logarithm.
logโกex \log_e x
It is written as:
lnโกx \ln x
where:
eโ‰ˆ2.71828 e\approx2.71828
๐ŸŸข 12. RECIPROCAL PROPERTY
logโกa(1x)=โˆ’logโกax \log_a\left(\frac{1}{x}\right) = -\log_a x
๐ŸŸข 13. LOGARITHM OF A RECIPROCAL
logโกa(1x)=logโกa(xโˆ’1) \log_a\left(\frac{1}{x}\right) = \log_a(x^{-1})
Therefore:
=โˆ’logโกax =-\log_a x
๐ŸŸข 14. CHANGE OF BASE BETWEEN TWO LOGARITHMS A useful identity is:
logโกab=1logโกba \log_a b = \frac{1}{\log_b a}
๐ŸŸข 15. PRODUCT OF LOGARITHMS
logโกabร—logโกbc=logโกac \log_a b\times\log_b c = \log_a c
More generally:
logโกabร—logโกbcร—logโกcd=logโกad \log_a b\times\log_b c\times\log_c d = \log_a d
๐ŸŸข 16. SPECIAL IDENTITY
logโกabร—logโกba=1 \log_a b\times\log_b a=1
๐ŸŸข 17. LOGARITHM OF 1 For any valid base:
logโกa1=0 \log_a1=0
๐ŸŸข 18. LOGARITHM OF THE BASE
logโกaa=1 \log_a a=1
๐ŸŸข 19. EXPONENTIAL FORM AND LOGARITHMIC FORM The two forms are equivalent. Exponential form:
ax=N a^x=N
Logarithmic form:
logโกaN=x \log_aN=x
๐ŸŸข 20. SOLVING A SIMPLE LOGARITHMIC EQUATION If:
logโกax=n \log_a x=n
then convert to exponential form:
x=an x=a^n
๐ŸŸข 21. SOLVING LOGARITHMIC EQUATIONS WITH THE SAME BASE If:
logโกax=logโกay \log_a x=\log_a y
then:
x=y x=y
provided x and y are positive. ๐ŸŸข 22. LOGARITHMIC EQUATION USING PRODUCT RULE If:
logโกax+logโกay \log_a x+\log_a y
then:
logโกa(xy) \log_a(xy)
can be used to combine the logarithms. ๐ŸŸข 23. LOGARITHMIC EQUATION USING QUOTIENT RULE If:
logโกaxโˆ’logโกay \log_a x-\log_a y
then:
logโกa(xy) \log_a\left(\frac{x}{y}\right)
can be used to combine the logarithms. ๐ŸŸข 24. LOGARITHMIC EQUATION USING POWER RULE If:
nlogโกax n\log_a x
then:
logโกa(xn) \log_a(x^n)
can be used to combine the expression. ๐ŸŸข 25. ANTILOGARITHM If:
logโกax=y \log_a x=y
then:
x=ay x=a^y
The value x is called the antilogarithm of y to base a. ๐ŸŸข 26. CHARACTERISTIC AND MANTISSA For common logarithms:
logโก10N \log_{10}N
the integer part is called the characteristic and the decimal part is called the mantissa. For example, if:
logโกN=2.3456 \log N=2.3456
then: Characteristic:
2 2
Mantissa:
0.3456 0.3456
๐ŸŸข 27. LOGARITHM OF NUMBERS BETWEEN 0 AND 1 If:
0<x<1 0<x<1
then:
logโก10x<0 \log_{10}x<0
For example:
logโก10(0.1)=โˆ’1 \log_{10}(0.1)=-1
๐ŸŸข 28. LOGARITHM AND EXPONENTS The following identities are important:
logโกa(ax)=x \log_a(a^x)=x
and:
alogโกax=x a^{\log_a x}=x
๐ŸŸก KEY POINTS 1. A logarithm is the inverse operation of exponentiation. 2. Always check the base and argument before applying logarithm rules. 3. The base must be positive and cannot be 1. 4. The argument of a real logarithm must be positive. 5. Product becomes addition:
logโกa(xy)=logโกax+logโกay \log_a(xy) = \log_a x+\log_a y
6. Quotient becomes subtraction:
logโกa(xy)=logโกaxโˆ’logโกay \log_a\left(\frac{x}{y}\right) = \log_a x-\log_a y
7. Power comes in front:
logโกa(xn)=nlogโกax \log_a(x^n) = n\log_a x
8. Remember:
logโกa1=0 \log_a1=0
9. Remember:
logโกaa=1 \log_aa=1
10. Convert logarithmic equations into exponential form whenever it makes solving easier. 11. Do not use:
logโกa(x+y)=logโกax+logโกay \log_a(x+y) = \log_ax+\log_ay
This is NOT a valid logarithm rule. 12. Similarly:
logโกa(xโˆ’y) \log_a(x-y)
cannot generally be separated into two logarithms. ๐Ÿ”ด EXAM TIP When you see a logarithm problem: 1. Check the base. 2. Check that every logarithm argument is positive. 3. Apply product, quotient or power rules. 4. Try to express logarithms with the same base. 5. Convert to exponential form when necessary. 6. Check the final answer against the domain restrictions.

Example

๐Ÿ”ต LOGARITHMS - SOLVED EXAMPLES ๐ŸŸ  EXAMPLE 1: BASIC LOGARITHM QUESTION: Find:
logโก28 \log_2 8
๐ŸŸข SOLUTION: Convert to exponential form:
2x=8 2^x=8
Since:
23=8 2^3=8
Therefore:
x=3 x=3
Hence:
logโก28=3 \log_2 8=3
โœ… ANSWER:
33
๐ŸŸ  EXAMPLE 2: LOGARITHM OF 1 QUESTION: Find:
logโก51 \log_5 1
๐ŸŸข SOLUTION: We know:
50=1 5^0=1
Therefore:
logโก51=0 \log_5 1=0
โœ… ANSWER:
00
๐ŸŸ  EXAMPLE 3: LOGARITHM OF THE BASE QUESTION: Find:
logโก77 \log_7 7
๐ŸŸข SOLUTION: Since:
71=7 7^1=7
Therefore:
logโก77=1 \log_7 7=1
โœ… ANSWER:
1 1
๐ŸŸ  EXAMPLE 4: LOGARITHM OF A POWER QUESTION: Find:
logโก381 \log_3 81
๐ŸŸข SOLUTION: Write 81 as a power of 3:
81=34 81=3^4
Therefore:
logโก381=logโก3(34) \log_3 81 = \log_3(3^4)
=4 =4
โœ… ANSWER:
44
๐ŸŸ  EXAMPLE 5: PRODUCT RULE QUESTION: Simplify:
logโก24+logโก28 \log_2 4+\log_2 8
๐ŸŸข SOLUTION: Using:
logโกax+logโกay=logโกa(xy) \log_a x+\log_a y = \log_a(xy)
Therefore:
logโก24+logโก28=logโก2(4ร—8) \log_2 4+\log_2 8 = \log_2(4\times8)
=logโก232 =\log_2 32
Since:
32=25 32=2^5
Therefore:
=5 =5
โœ… ANSWER:
55
๐ŸŸ  EXAMPLE 6: QUOTIENT RULE QUESTION: Simplify:
logโก381โˆ’logโก39 \log_3 81-\log_3 9
๐ŸŸข SOLUTION: Using:
logโกaxโˆ’logโกay=logโกa(xy) \log_a x-\log_a y = \log_a\left(\frac{x}{y}\right)
Therefore:
logโก381โˆ’logโก39=logโก3(819) \log_3 81-\log_3 9 = \log_3\left(\frac{81}{9}\right)
=logโก39 =\log_3 9
=2 =2
โœ… ANSWER:
22
๐ŸŸ  EXAMPLE 7: POWER RULE QUESTION: Simplify:
3logโก24 3\log_2 4
๐ŸŸข SOLUTION: Using:
nlogโกax=logโกa(xn) n\log_a x = \log_a(x^n)
Therefore:
3logโก24=logโก2(43) 3\log_2 4 = \log_2(4^3)
=logโก264 =\log_2 64
=6 =6
โœ… ANSWER:
66
๐ŸŸ  EXAMPLE 8: SOLVING A SIMPLE LOGARITHMIC EQUATION QUESTION: Solve:
logโก2x=5 \log_2 x=5
๐ŸŸข SOLUTION: Convert to exponential form:
x=25 x=2^5
x=32 x=32
โœ… ANSWER:
x=32x = 32
๐ŸŸ  EXAMPLE 9: SOLVING WITH A DIFFERENT BASE QUESTION: Solve:
logโก5x=3 \log_5 x=3
๐ŸŸข SOLUTION: Convert to exponential form:
x=53 x=5^3
x=125 x=125
โœ… ANSWER:
x=125x = 125
๐ŸŸ  EXAMPLE 10: SOLVING A LOGARITHMIC EQUATION QUESTION: Solve:
logโก3(x)=logโก3(27) \log_3(x)=\log_3(27)
๐ŸŸข SOLUTION: Since the bases are the same:
x=27 x=27
Therefore:
x=27 x=27
โœ… ANSWER:
x=27x = 27
๐ŸŸ  EXAMPLE 11: SUM OF LOGARITHMS QUESTION: Solve:
logโก2x+logโก24=5 \log_2 x+\log_2 4=5
๐ŸŸข SOLUTION: Using the product rule:
logโก2(4x)=5 \log_2(4x)=5
Convert to exponential form:
4x=25 4x=2^5
4x=32 4x=32
x=8 x=8
โœ… ANSWER:
x=8x = 8
๐ŸŸ  EXAMPLE 12: DIFFERENCE OF LOGARITHMS QUESTION: Solve:
logโก3xโˆ’logโก32=2 \log_3 x-\log_3 2=2
๐ŸŸข SOLUTION: Using the quotient rule:
logโก3(x2)=2 \log_3\left(\frac{x}{2}\right)=2
Convert to exponential form:
x2=32 \frac{x}{2}=3^2
x2=9 \frac{x}{2}=9
x=18 x=18
โœ… ANSWER:
x=18x = 18
๐ŸŸ  EXAMPLE 13: LOGARITHM WITH A POWER QUESTION: Find:
logโก5(252) \log_5(25^2)
๐ŸŸข SOLUTION: Since:
25=52 25=5^2
Therefore:
252=(52)2 25^2=(5^2)^2
=54 =5^4
Hence:
logโก5(54)=4 \log_5(5^4)=4
โœ… ANSWER:
44
๐ŸŸ  EXAMPLE 14: CHANGE OF BASE QUESTION: Express:
logโก27 \log_2 7
using common logarithms. ๐ŸŸข SOLUTION: Using the change of base formula:
logโกax=logโกxlogโกa \log_a x = \frac{\log x}{\log a}
Therefore:
logโก27=logโก7logโก2 \log_2 7 = \frac{\log 7}{\log 2}
โœ… ANSWER:
logโก7logโก2 \frac{\log 7}{\log 2}
๐ŸŸ  EXAMPLE 15: RECIPROCAL PROPERTY QUESTION: Simplify:
logโก2(18) \log_2\left(\frac{1}{8}\right)
๐ŸŸข SOLUTION: Write:
18=2โˆ’3 \frac{1}{8}=2^{-3}
Therefore:
logโก2(2โˆ’3)=โˆ’3 \log_2(2^{-3}) = -3
โœ… ANSWER:
โˆ’3-3
๐ŸŸ  EXAMPLE 16: PRODUCT OF LOGARITHMS QUESTION: Find:
logโก23ร—logโก38 \log_2 3\times\log_3 8
๐ŸŸข SOLUTION: Using:
logโกabร—logโกbc=logโกac \log_a b\times\log_b c = \log_a c
Therefore:
logโก23ร—logโก38=logโก28 \log_2 3\times\log_3 8 = \log_2 8
Since:
8=23 8=2^3
Therefore:
logโก28=3 \log_2 8=3
โœ… ANSWER:
33
๐ŸŸ  EXAMPLE 17: SPECIAL IDENTITY QUESTION: Find:
logโก25ร—logโก52 \log_2 5\times\log_5 2
๐ŸŸข SOLUTION: Using:
logโกabร—logโกba=1 \log_a b\times\log_b a=1
Therefore:
logโก25ร—logโก52=1 \log_2 5\times\log_5 2=1
โœ… ANSWER:
11
๐ŸŸ  EXAMPLE 18: SOLVING USING PRODUCT RULE QUESTION: Solve:
logโก2x+logโก2(xโˆ’2)=3 \log_2 x+\log_2(x-2)=3
๐ŸŸข SOLUTION: First combine the logarithms:
logโก2[x(xโˆ’2)]=3 \log_2[x(x-2)]=3
Convert to exponential form:
x(xโˆ’2)=23 x(x-2)=2^3
x2โˆ’2x=8 x^2-2x=8
x2โˆ’2xโˆ’8=0 x^2-2x-8=0
Factor:
(xโˆ’4)(x+2)=0 (x-4)(x+2)=0
Therefore:
x=4 x=4
or:
x=โˆ’2 x=-2
Since logarithm arguments must be positive:
x>2 x>2
Therefore:
x=4 x=4
โœ… ANSWER:
x=4x = 4
๐ŸŸ  EXAMPLE 19: SOLVING USING QUOTIENT RULE QUESTION: Solve:
logโก3(x)โˆ’logโก3(xโˆ’2)=1 \log_3(x)-\log_3(x-2)=1
๐ŸŸข SOLUTION: Using the quotient rule:
logโก3(xxโˆ’2)=1 \log_3\left(\frac{x}{x-2}\right)=1
Convert to exponential form:
xxโˆ’2=3 \frac{x}{x-2}=3
Therefore:
x=3(xโˆ’2) x=3(x-2)
x=3xโˆ’6 x=3x-6
2x=6 2x=6
x=3 x=3
Check:
xโˆ’2=1>0 x-2=1>0
Therefore the solution is valid. โœ… ANSWER:
x=3 x = 3
๐ŸŸ  EXAMPLE 20: SOLVING A LOGARITHMIC EQUATION QUESTION: Solve:
logโก2(x+1)=3 \log_2(x+1)=3
๐ŸŸข SOLUTION: Convert to exponential form:
x+1=23 x+1=2^3
x+1=8 x+1=8
x=7 x=7
โœ… ANSWER:
x=7x = 7
๐ŸŸ  EXAMPLE 21: LOGARITHM WITH COEFFICIENT QUESTION: Solve:
2logโก3x=4 2\log_3 x=4
๐ŸŸข SOLUTION: Divide both sides by 2:
logโก3x=2 \log_3 x=2
Convert to exponential form:
x=32 x=3^2
x=9 x=9
โœ… ANSWER:
x=9x = 9
๐ŸŸ  EXAMPLE 22: COMBINING MULTIPLE LOGARITHMS QUESTION: Simplify:
logโก28+logโก24โˆ’logโก22 \log_2 8+\log_2 4-\log_2 2
๐ŸŸข SOLUTION: Using the product and quotient rules:
=logโก2(8ร—42) = \log_2\left(\frac{8\times4}{2}\right)
=logโก216 =\log_2 16
Since:
16=24 16=2^4
Therefore:
=4 =4
โœ… ANSWER:
44
๐ŸŸ  EXAMPLE 23: EXPONENTIAL FORM TO LOGARITHMIC FORM QUESTION: Express:
34=81 3^4=81
in logarithmic form. ๐ŸŸข SOLUTION: Using:
ax=Nโ€…โ€ŠโŸบโ€…โ€ŠlogโกaN=x a^x=N \iff \log_aN=x
Therefore:
logโก381=4 \log_3 81=4
โœ… ANSWER:
logโก381=4 \log_3 81=4
๐ŸŸ  EXAMPLE 24: LOGARITHMIC FORM TO EXPONENTIAL FORM QUESTION: Express:
logโก5125=3 \log_5 125=3
in exponential form. ๐ŸŸข SOLUTION: Using:
logโกaN=xโ€…โ€ŠโŸบโ€…โ€Šax=N \log_aN=x \iff a^x=N
Therefore:
53=125 5^3=125
โœ… ANSWER:
53=125 5^3=125
๐ŸŸ  EXAMPLE 25: CHANGE OF BASE TO NATURAL LOGARITHMS QUESTION: Express:
logโก47 \log_4 7
using natural logarithms. ๐ŸŸข SOLUTION: Using:
logโกax=lnโกxlnโกa \log_a x = \frac{\ln x}{\ln a}
Therefore:
logโก47=lnโก7lnโก4 \log_4 7 = \frac{\ln7}{\ln4}
โœ… ANSWER:
lnโก7lnโก4 \frac{\ln7}{\ln4}