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Logarithms Concepts
QUANTITATIVEAPTITUDE

Logarithms Concepts

Learn logarithm laws, bases, transformations, and techniques for solving aptitude problems.

1. Basic Logarithm Properties & Simplification

  • Formula:
loga(MN)=logaM+logaN \log_a(MN)=\log_aM+\log_aN loga(MN)=logaMlogaN \log_a\left(\frac MN\right)=\log_aM-\log_aN loga(Mn)=nlogaM \log_a(M^n)=n\log_aM logaa=1,loga1=0 \log_a a=1,\qquad \log_a1=0

Use these to expand or combine logarithmic expressions.

  • Example: Simplify (\log(8\times125)). Solution:
log(8×125)=log8+log125\log(8\times125)=\log8+\log125 =log23+log53=\log2^3+\log5^3 =3log2+3log5=3\log2+3\log5 =3log(10)=3\log(10) 3\boxed{3}

2. Combining Logarithms

  • Formula:
logaM+logaN=loga(MN) \log_aM+\log_aN=\log_a(MN) logaMlogaN=loga(MN) \log_aM-\log_aN=\log_a\left(\frac MN\right)

Use when several logarithms with the same base need to be combined.

  • Example: Simplify
log212log23+log24. \log_2 12-\log_2 3+\log_2 4.

Solution:

=log2(12×43)=\log_2\left(\frac{12\times4}{3}\right) =log216=\log_2 16 4\boxed{4}

3. Change of Base

  • Formula:
logab=logcblogca \boxed{\log_a b=\frac{\log_c b}{\log_c a}}

In particular:

logab=lnblna \log_a b=\frac{\ln b}{\ln a}

Use when the required base is inconvenient or different logarithms have different bases.

  • Example: Find (\log_2 8) using common logarithms. Solution:
log28=log108log102\log_2 8=\frac{\log10 8}{\log10 2}

Since:

8=238=2^3 log28=3\log_2 8=\boxed{3}

4. Converting Between Logarithmic and Exponential Forms

  • Formula:
logax=y    ay=x \boxed{\log_a x=y\iff a^y=x}

where (a>0,\ a\neq1,\ x>0).

  • Example: Solve
log3x=4. \log_3x=4.

Solution:

Convert to exponential form:

x=34x=3^4 x=81\boxed{x=81}

5. Basic Logarithmic Equations

  • Formula:
logaf(x)=logag(x)f(x)=g(x) \log_a f(x)=\log_a g(x) \Rightarrow f(x)=g(x)

provided both arguments are positive.

  • Example: Solve
log2(x+3)=log2(2x1). \log_2(x+3)=\log_2(2x-1).

Solution:

Equate the arguments:

x+3=2x1x+3=2x-1 x=4x=4

Check:

x+3=7>0,2x1=7>0x+3=7>0,\qquad2x-1=7>0

Therefore:

x=4\boxed{x=4}

6. Logarithmic Equations with Different Bases

  • Formula:
logax=logbxlogba \log_a x=\frac{\log_bx}{\log_ba}

Convert all logarithms to a common base.

  • Example: Solve
log2x+log4x=6. \log_2x+\log_4x=6.

Solution:

Let:

y=log2xy=\log_2x

Since:

log4x=log2xlog24=y2\log_4x=\frac{\log_2x}{\log_24} =\frac y2

Therefore:

y+y2=6y+\frac y2=6 3y2=6\frac{3y}{2}=6 y=4y=4

Hence:

x=24x=2^4 x=16\boxed{x=16}

7. Logarithmic Equations Using Product/Quotient Rules

  • Formula:
logaM+logaN=loga(MN) \log_aM+\log_aN=\log_a(MN) logaMlogaN=loga(M/N) \log_aM-\log_aN=\log_a(M/N)

Combine first, then convert to exponential form.

  • Example: Solve
log3x+log3(x2)=1. \log_3x+\log_3(x-2)=1.

Solution:

Domain:

x>0,x2>0x>0,\quad x-2>0

Thus:

x>2x>2

Combine:

log3[x(x2)]=1\log_3[x(x-2)]=1

Therefore:

x(x2)=3x(x-2)=3 x22x3=0x^2-2x-3=0 (x3)(x+1)=0(x-3)(x+1)=0 x=3,1x=3,-1

Only (x=3) satisfies (x>2).

x=3\boxed{x=3}

8. Characteristic and Mantissa

  • Formula:
log10N=Characteristic+Mantissa \log_{10}N=\text{Characteristic}+\text{Mantissa}

For (N>1):

Characteristic=number of digits1 \boxed{\text{Characteristic}=\text{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\log(2^{50})=50\log2 =50(0.3010)=15.05=50(0.3010)=15.05

Therefore:

Characteristic=15\text{Characteristic}=15

Number of digits:

15+1=1615+1=\boxed{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 \boxed{\text{Number of digits}=\lfloor\log_{10}N\rfloor+1}

For (a^n):

digits=nlog10a+1 \boxed{\text{digits}=\lfloor n\log_{10}a\rfloor+1}
  • Example: How many digits are there in (3^{20}), given (\log3=0.4771)? Solution:
log(320)=20(0.4771)=9.542\log(3^{20})=20(0.4771)=9.542

Therefore:

digits=9.542+1\text{digits}=\lfloor9.542\rfloor+1 =9+1=9+1 10\boxed{10}

10. Logarithmic Inequalities — Base Greater Than 1

  • Formula: If (a>1):
logax>logay    x>y \boxed{\log_a x>\log_a y\iff x>y}

If:

logaf(x)>k \log_a f(x)>k

then:

f(x)>ak f(x)>a^k

Always impose (f(x)>0).

  • Example: Solve
log2(x1)>3. \log_2(x-1)>3.

Solution:

Since (2>1):

x1>23x-1>2^3 x1>8x-1>8 x>9\boxed{x>9}

The domain (x>1) is automatically satisfied.


11. Logarithmic Inequalities — Base Between 0 and 1

  • Formula: If:
0<a<1, 0<a<1,

the inequality reverses:

logax>logay    x<y \boxed{\log_a x>\log_a y\iff x<y}

Always require (x>0).

  • Example: Solve
log0.5(x1)>2. \log_{0.5}(x-1)>-2.

Solution:

Since:

0<0.5<1,0<0.5<1,

the inequality reverses:

x1<(0.5)2x-1<(0.5)^{-2} x1<4x-1<4 x<5x<5

Domain:

x1>0x>1x-1>0\Rightarrow x>1

Therefore:

1<x<5\boxed{1<x<5}

12. Domain Restrictions in Logarithms

  • Formula: For every logarithm:
argument>0 \boxed{\text{argument}>0}

and:

base>0,base1 \boxed{\text{base}>0,\quad\text{base}\neq1}

Apply these conditions after solving the equation.

  • Example: Solve
log2(x29)=2. \log_2(x^2-9)=2.

Solution:

Convert to exponential form:

x29=22x^2-9=2^2 x2=13x^2=13 x=±13x=\pm\sqrt{13}

Both values give:

x29=4>0x^2-9=4>0

Therefore:

x=±13\boxed{x=\pm\sqrt{13}}

13. Logarithm of a Power

  • Formula:
loga(xn)=nlogax \boxed{\log_a(x^n)=n\log_ax}

Also:

logamx=1mlogax \boxed{\log_{a^m}x=\frac1m\log_ax}

Use to reduce powers or convert between bases.

  • Example: Simplify
log2(323). \log_2(32^3).

Solution:

32=2532=2^5

Therefore:

log2(323)============log2(215)\log_2(32^3) ============ \log_2(2^{15}) 15\boxed{15}

14. Reciprocal and Negative Logarithms

  • Formula:
loga1x=logax \boxed{\log_a\frac1x=-\log_ax} log1/ax=logax \boxed{\log_{1/a}x=-\log_ax}

Use when reciprocals or reciprocal bases appear.

  • Example: Simplify
log218. \log_2\frac18.

Solution:

18=23\frac18=2^{-3}

Therefore:

log2(23)=3\log_2(2^{-3})=-3 3\boxed{-3}

15. Exponential Equations Using Logarithms

  • Formula: For:
ax=b, a^x=b, x=logbloga \boxed{x=\frac{\log b}{\log a}}

Use when the equation cannot be solved conveniently by matching powers.

  • Example: Solve
5x=20. 5^x=20.

Solution:

Take logarithms:

xlog5=log20x\log5=\log20

Therefore:

x=log20log5x=\frac{\log20}{\log5} x=log520\boxed{x=\log_5 20}

Advanced Variants

16. Nested Logarithms

  • Formula:
loga(logbx)=clogbx=acx=bac \boxed{\log_a(\log_bx)=c \Rightarrow \log_bx=a^c \Rightarrow x=b^{a^c}}

Domain restrictions must be checked at every level.

  • Example: Solve
log2(log3x)=2. \log_2(\log_3x)=2.

Solution:

log3x=22=4\log_3x=2^2=4

Therefore:

x=34x=3^4 x=81\boxed{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(x1)+log2(x+1)=3. \log_2(x-1)+\log_2(x+1)=3.

Solution:

Domain:

x>1x>1

Combine:

log2[(x1)(x+1)]=3\log_2[(x-1)(x+1)]=3 x21=8x^2-1=8 x2=9x^2=9 x=±3x=\pm3

Since (x>1):

x=3\boxed{x=3}

18. Logarithmic Inequality with Multiple Terms

  • Formula: Combine logarithms using:
logaM+logaN=loga(MN) \log_aM+\log_aN=\log_a(MN)

and remember that the inequality reverses when (0<a<1).

  • Example: Solve
log2x+log2(x2)>3. \log_2x+\log_2(x-2)>3.

Solution:

Domain:

x>2x>2

Combine:

log2[x(x2)]>3\log_2[x(x-2)]>3

Since (2>1):

x(x2)>8x(x-2)>8 x22x8>0x^2-2x-8>0 (x4)(x+2)>0(x-4)(x+2)>0

Thus:

x<2orx>4x<-2\quad\text{or}\quad x>4

Applying (x>2):

x>4\boxed{x>4}

19. Change-of-Base Product/Cyclic Form

  • Formula:
logablogbc=logac \boxed{\log_ab\cdot\log_bc=\log_ac}

and:

logablogbclogca=1 \boxed{\log_ab\cdot\log_bc\cdot\log_ca=1}

Useful for aptitude questions involving cyclic logarithms.

  • Example: Evaluate
log23log35log58. \log_2 3\cdot\log_3 5\cdot\log_5 8.

Solution:

Using:

logablogbc=logac\log_ab\cdot\log_bc=\log_ac log23log35=log25\log_2 3\cdot\log_3 5=\log_2 5

Therefore:

log25log58=log28\log_2 5\cdot\log_5 8 =\log_2 8 3\boxed{3}

20. Base Conditions in Logarithmic Inequalities

  • Formula: For:
logax \log_a x

if (a>1), the function is increasing; if (0<a<1), it is decreasing.

a>1: preserve inequality \boxed{a>1:\text{ preserve inequality}} 0<a<1: reverse inequality \boxed{0<a<1:\text{ reverse inequality}}
  • Example: Solve
log1/3(2x1)2. \log_{1/3}(2x-1)\leq2.

Solution:

Since:

0<13<1,0<\frac13<1,

reverse the inequality:

2x1(13)22x-1\geq\left(\frac13\right)^2 2x1192x-1\geq\frac19 2x1092x\geq\frac{10}{9} x59x\geq\frac59

Domain:

2x1>0x>122x-1>0\Rightarrow x>\frac12

Combining:

x59\boxed{x\geq\frac59}

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