Logarithms

• 1: Definition and properties
• 2: Proofs of Logarithmic properties
• 3: Practice Problems on logarithms

1 - Definition and properties

Logarithm: For a number and base, the power the base should have so that it is equal to the given number is called a logarithm.For example, the logarithm of 8 with base 2 means what power should 2 have so that it is equal to 8. 2³=8, the answer is 3.

logarithm is represented as log_b a where a and b are non-zero positive numbers.

Where a is the number inside the logarithmic function and b is the base. And we read it as log a base b.

Symbolically the above example can be represented as: Log₂ 8 =3

I) find the values for the following logarithms:

1. log₃ 27
2. log₁₀ 100
3. log₄ 2
4. log₁₀ (0.1)

Solutions are at the end of the article.

Logarithm in fact is an inverse of exponential function. Let’s see how logarithm and exponents are interconvertible:

a^b=x ------> b= log_a x

Look at the example 2³=8, this is represented as log₂ 8=3

This is how we interconvert exponents into logarithms and vice versa.

II) convert the following logarithms into exponents

1. log_a b = x
2. log₂₅ 5= 1/2
3. log₃ 9= 2

Solutions are at the end of the article.

Properties:

For a, b, x, y being postive non-zero numbers.

Product rule: Log_a xy = log_a x+ log_a y

Quotient rule: Log_a $\frac{x}{y}$ = log_a x- log_a y

Power rule: (i) log_b $a^m$ = m log_b a

Power rule: (ii) log_bⁿ a= $\frac{1}{n}$ log_b a

log_b a= $\frac{1}{log_a b}$

log_b a= $\frac{log_x a}{log_x b}$

log_a 1 = 0

Other properties:

log_a a= 1

If log_b a= log_b c then a=c

log_b $\frac{1}{a}$=log_{$\frac{1}{b}$} a= -log_a a

a^{log_a x} = x

log_a 0 is undefined

III) Solve the following logarithms:

1. log₅ x = log₅ 2 + log₅ 7
2. log_b x+log_b 1/x
3. log₅ 15 - log₅ 3
4. log₅ 125
5. log₁₆ 4

Solutions are at the end of the article.

Sometimes we drop bases in logarithms for base 10 and natural logarithms. Log₁₀ a can be represented dropping 10 i.e. log a. And logarithm with natural base ’e’ can be represented as ln x (=log_e x)

Solutions:

I

1) ans=3

Sol: log₃ 27 = log₃ $3^3$

Using the property $log_b$ $a^n$ = n $log_b$ a

= 3log₃ 3

Using the property log_a a=1

= 3

2) ans=2

Sol:

log₁₀ 100 = log₁₀ $10^2$

Using the property log_b $a^n$ = n log_b a

= 2log₁₀ 10

Using the property log_a a=1

=2

3) ans=½

Sol:

log₄ 2 = log_2² 2

Using the property log_bⁿ a = $\frac{1}{n}$ log_b a

= $\frac{1}{2}$ log₂ 2

Using the property log_a a=1

= $\frac{1}{2}$

4) ans=-1

Sol:

Log₁₀ (0.1) = log₁₀ $\frac{1}{10}$

Using the property Log_a $\frac{x}{y}$ =log_a x- log_a y

= - log₁₀ 10

Using the property log_a a=1 =-1

II)

1. $a^x$
2. 5= 25^{$\frac{1}{2}$}
3. 9= $3^2$

III)

1) x=14

Sol:

log₅ x = log₅ 2 + log₅ 7

Using the property Log_a xy=log_a x+ log_a y

log₅ x = log₅ 2*7

Using the property if log_a b=log_a c then b=c

x=14

2) ans=0

Sol:

log_b x+log_b 1/x

Using the property Log_a xy =log_a x+ log_a y

=log_b x* $\frac{1}{x}$

=log_b 1

Using the property Log_a 1 = 0

=0

3) ans= 0

Sol:

log₅ 15 - log₅ 3

Using the property Log_a $\frac{x}{y}$ =log_a x- log_a y

=Log₅ $\frac{15}{3}$

= log₅ 5

Using the property Log_a 1 = 0

=0

4) ans =3

Sol:

log₅ 125

=log₅ $5^3$

Using the property log_b aⁿ = n log_b a

=3 log₅ 5

Using the property log_a a=1

=3

5) ans =½

Sol:

log₁₆ 4

=log_4² 4

Using the property log_bⁿ a = $\frac{1}{n}$ log_b a

=½ log₄ 4

Using the property log_a a=1

=½

2 - Proofs of Logarithmic properties

1) Product Rule: Log_a xy = log_a x+ log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

multiply the above equations

xy=a^m * aⁿ

Using the property b^p * b^q=b^p+q from exponents we can write the above equation as follows

xy=a^m+n

Convert back into logarithm

log_a xy = m+n

on substituting the values m= log_a x and n =log_a y we get

log_a xy = log_a x + log_a y

2) Quotient Rule: Log_a $\frac{x}{y}$ =log_a x- log_a y

proof:

let log_a x = m, log_a y = n

convert these logarithms into exponents

x=a^m and y=aⁿ

divide the above equations

$\frac{x}{y}$ = $\frac{a^m}{a^n}$

using the property b^p/b^q=b^p-q from exponents we can write the above equation as follows

$\frac{x}{y}$=a^m-n

convert into logarithm

log_a $\frac{x}{y}$=m-n

on substituting the m= log_a x & n =log_a y we get

log_a (xy)=log_a x - log_a y

3) Power Rules:

i) log_b $a^m$ = m log_b a

Proof:

let x=log_b a

convert logarithm into exponent

b^x=a

taking both sides of the equation to m’th power

(b^x)^m=a^m

using the property (a^m)ⁿ=a^mn from exponents we can write the above equation as follows

b^xm=a^m

convert back to logarithm

xm=log_b a^m

substitute x=log_b a back in the equation and interchanging

log_b a^m = m log_b a

ii) log_bⁿ a= $\frac{1}{n}$ log_b a

proof:

let log_bⁿ a=x

convert this into exponential form

a=(bⁿ)^x

use exponential property

(a^m)ⁿ=a^mn=(aⁿ)^m

a=(b^x)ⁿ

use the property if a^m=b then a=b^{$\frac{1}{m}$} from exponents we can write the above equation as follows

a^{$\frac{1}{n}$}=b^x

interchanging and converting into logarithmic form

b^x=a^{$\frac{1}{n}$}

x=log^b a^{$\frac{1}{n}$}

substitute the value of x=log_bⁿ a and use logarithmic property log_b a^m= m log_b a

log_bⁿ a= $\frac{1}{n}$ log_b a

4) log of 1 rule: log_b 1=0

Proof:

from exponents we have:

b⁰=1

convert this into logarithmic form

0=_b 1

5) Change of Base Rule: log_b a= $\frac{log_x a}{log_x b}$

Proof:

let log_x a=p, log_x b=q

Convert these logarithms into exponential forms

a=x^p and b=x^q

Substituting in the Left hand side of the to solve into right hand side of the Rule

L.H.S. : log_b a= log_xⁿ x^m

Using the power rule: log_n^q a^p= $\frac{p}{q}$ $log_b$ a

log_b a= $\frac{p}{q}$ log_x x

Using the Rule log_a a=1

Log_b a= $\frac{p}{q}$

Substituting the assumed p= $log_x$ a, q= $log_x$ b

Log_b a= $\frac{log_x a}{log_x b}$

6) log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

Proof:

i) Log_b (1/a)=log_b a^-1

Using the property log_a b^m=m log_a b =-log_b a

ii) log_{$\frac{1}{b}$} a=log_$b^{-1}$ a

log_b $\frac{1}{a}$ =log_{$\frac{1}{b}$} a= -log_b a

7) a^{log_a x} = x

Proof:

Let p=a^{log_a x}

Converting the above exponent into logarithm

Log_a p = log_a x

Using he property if log_b a= log_b c then a=c

p=x

But we assumed p=a^{log_a x}

Therefore a^{log_a x} = x

3 - Practice Problems on logarithms

I) write the following in the logarithmic format

1. $5^3$ = 125

2. $2^4$ = 16

3. $4^{-2}=\frac{1}{16}$

4. $(\frac{1}{2})^{-3}=8$

5. $\sqrt {81}$ = 9

6. $x^{2a}=y$

II) convert the following logarithms into exponents

1. x= $log_3\ y$

2. $log_x\ 1$ =0

3. ln x = 5

4. $log\ 100=2$

5. $log_8\ 2=\frac{1}{3}$

III) Find the values of x

1. $log_5\ 25$ = x

2. $log_{100}\ 1000$ = x

3. $log_{23}\ 1$ = x

4. $log_3\ \frac{1}{27}$ = x

5. $log_x\ 49$ = 7

6. $log_9\ x$ = -3

7. $log_5\ \frac{1}{5}$ = x

8. $log_8\ x=1- \frac{1}{2}$

9. $log_{12}\ 1$ = x

10. $log_{32}\ 2$ = x

11. $log_4\ \frac{1}{32}$ = x

12. $log_{12}\ \frac{1}{144}$ = x

IV) Write the following expressions in terms of logs of x, y and z

1. log $x^3y$

2. log $xyz$

3. log $\sqrt[3]x y^2 z$

4. log $\frac{x}{z^3}y$

5. log $\sqrt[5]{x^3yz^2}$

6. log $x \sqrt{ \frac{\sqrt{x^3y^2}}{z^4}}$

7. log $(x^3y)^{\frac{1}{2}}$

8. log $\frac{x^3}{y}$

V) Solve the following logarithmic equations

1. ln x =-2

2. log (7x+2) = 2

3. log x + log (x+1) = log 4x

4. 2log x = log 2 + log(3x-4)

5. $log_3\ (x+3) + log_3\ (x+2) - log_3\ 10 = log_3\ x$

6. $log_2\ x + log_2\ (x+3) = 1$

VI) If log 2= x, log 3 =y then express the following in x and y

1. log 24

2. log 225

3. log 150

4. $log_7\ 980$

5. log 343

6. log 12.5

7. log 0.2

8. log 15

9. log 2.5

10. $log_7\ 3.5$

VII) Solve the following equations

1. $3^x$ - 1 =8
2. $2^{2x}-2^x-6$=0
3. $3^{1-x}=5^x$
4. $3^{2x-1}+3{x+2}-18$=0
5. $e^{2x}-2e^{x}-15$=0