1. Basic Logarithm Properties & Simplification
loga(MN)=logaM+logaN
loga(NM)=logaM−logaN
loga(Mn)=nlogaM
logaa=1,loga1=0
Use these to expand or combine logarithmic expressions.
- Example: Simplify (\log(8\times125)).
Solution:
log(8×125)=log8+log125
=log23+log53
=3log2+3log5
=3log(10)
3
2. Combining Logarithms
logaM+logaN=loga(MN)
logaM−logaN=loga(NM)
Use when several logarithms with the same base need to be combined.
log212−log23+log24.
Solution:
=log2(312×4)
=log216
4
3. Change of Base
logab=logcalogcb
In particular:
logab=lnalnb
Use when the required base is inconvenient or different logarithms have different bases.
- Example: Find (\log_2 8) using common logarithms.
Solution:
log28=log102log108
Since:
8=23
log28=3
logax=y⟺ay=x
where (a>0,\ a\neq1,\ x>0).
log3x=4.
Solution:
Convert to exponential form:
x=34
x=81
5. Basic Logarithmic Equations
logaf(x)=logag(x)⇒f(x)=g(x)
provided both arguments are positive.
log2(x+3)=log2(2x−1).
Solution:
Equate the arguments:
x+3=2x−1
x=4
Check:
x+3=7>0,2x−1=7>0
Therefore:
x=4
6. Logarithmic Equations with Different Bases
logax=logbalogbx
Convert all logarithms to a common base.
log2x+log4x=6.
Solution:
Let:
y=log2x
Since:
log4x=log24log2x=2y
Therefore:
y+2y=6
23y=6
y=4
Hence:
x=24
x=16
7. Logarithmic Equations Using Product/Quotient Rules
logaM+logaN=loga(MN)
logaM−logaN=loga(M/N)
Combine first, then convert to exponential form.
log3x+log3(x−2)=1.
Solution:
Domain:
x>0,x−2>0
Thus:
x>2
Combine:
log3[x(x−2)]=1
Therefore:
x(x−2)=3
x2−2x−3=0
(x−3)(x+1)=0
x=3,−1
Only (x=3) satisfies (x>2).
x=3
8. Characteristic and Mantissa
log10N=Characteristic+Mantissa
For (N>1):
Characteristic=number of digits−1
If (0<N<1), the characteristic is negative and is handled using bar notation.
- Example: Given (\log10 2=0.3010), find the number of digits in (2^{50}).
Solution:
log(250)=50log2
=50(0.3010)=15.05
Therefore:
Characteristic=15
Number of digits:
15+1=16
9. Number of Digits Using Logarithms
- Formula:
If (N>0) and (N) is not an exact power of 10:
Number of digits=⌊log10N⌋+1
For (a^n):
digits=⌊nlog10a⌋+1
- Example: How many digits are there in (3^{20}), given (\log3=0.4771)?
Solution:
log(320)=20(0.4771)=9.542
Therefore:
digits=⌊9.542⌋+1
=9+1
10
10. Logarithmic Inequalities — Base Greater Than 1
logax>logay⟺x>y
If:
logaf(x)>k
then:
f(x)>ak
Always impose (f(x)>0).
log2(x−1)>3.
Solution:
Since (2>1):
x−1>23
x−1>8
x>9
The domain (x>1) is automatically satisfied.
11. Logarithmic Inequalities — Base Between 0 and 1
0<a<1,
the inequality reverses:
logax>logay⟺x<y
Always require (x>0).
log0.5(x−1)>−2.
Solution:
Since:
0<0.5<1,
the inequality reverses:
x−1<(0.5)−2
x−1<4
x<5
Domain:
x−1>0⇒x>1
Therefore:
1<x<5
12. Domain Restrictions in Logarithms
- Formula:
For every logarithm:
argument>0
and:
base>0,base=1
Apply these conditions after solving the equation.
log2(x2−9)=2.
Solution:
Convert to exponential form:
x2−9=22
x2=13
x=±13
Both values give:
x2−9=4>0
Therefore:
x=±13
13. Logarithm of a Power
loga(xn)=nlogax
Also:
logamx=m1logax
Use to reduce powers or convert between bases.
log2(323).
Solution:
32=25
Therefore:
log2(323)============log2(215)
15
14. Reciprocal and Negative Logarithms
logax1=−logax
log1/ax=−logax
Use when reciprocals or reciprocal bases appear.
log281.
Solution:
81=2−3
Therefore:
log2(2−3)=−3
−3
15. Exponential Equations Using Logarithms
ax=b,
x=logalogb
Use when the equation cannot be solved conveniently by matching powers.
5x=20.
Solution:
Take logarithms:
xlog5=log20
Therefore:
x=log5log20
x=log520
Advanced Variants
16. Nested Logarithms
loga(logbx)=c⇒logbx=ac⇒x=bac
Domain restrictions must be checked at every level.
log2(log3x)=2.
Solution:
log3x=22=4
Therefore:
x=34
x=81
17. Logarithmic Equations with Parameter
- Formula:
Combine logarithms first, convert to an algebraic equation, then apply the logarithm domain restrictions.
- Example: Solve
log2(x−1)+log2(x+1)=3.
Solution:
Domain:
x>1
Combine:
log2[(x−1)(x+1)]=3
x2−1=8
x2=9
x=±3
Since (x>1):
x=3
18. Logarithmic Inequality with Multiple Terms
- Formula:
Combine logarithms using:
logaM+logaN=loga(MN)
and remember that the inequality reverses when (0<a<1).
log2x+log2(x−2)>3.
Solution:
Domain:
x>2
Combine:
log2[x(x−2)]>3
Since (2>1):
x(x−2)>8
x2−2x−8>0
(x−4)(x+2)>0
Thus:
x<−2orx>4
Applying (x>2):
x>4
logab⋅logbc=logac
and:
logab⋅logbc⋅logca=1
Useful for aptitude questions involving cyclic logarithms.
log23⋅log35⋅log58.
Solution:
Using:
logab⋅logbc=logac
log23⋅log35=log25
Therefore:
log25⋅log58=log28
3
20. Base Conditions in Logarithmic Inequalities
logax
if (a>1), the function is increasing; if (0<a<1), it is decreasing.
a>1: preserve inequality
0<a<1: reverse inequality
log1/3(2x−1)≤2.
Solution:
Since:
0<31<1,
reverse the inequality:
2x−1≥(31)2
2x−1≥91
2x≥910
x≥95
Domain:
2x−1>0⇒x>21
Combining:
x≥95