INTRODUCTION TO LOGARITHM

Definition of Logarithm of a Number

The logarithm of a number x to the base b is the exponent y to which b is raised to obtain x.
In symbols,
Logarithms are exponents

𝐥 = if and only if =
argument / answer

base

for all positive real numbers x and b, and b ≠ 1.

The notation log is read as “the logarithm (or log) of x to the base b” or “the logarithm (or log)
base b of x”. The definition of a logarithm indicates that a logarithm is an exponent.
log = is the logarithmic form of =
= is the exponential form of log =

Key Concepts:
The inverse of an exponential function is a logarithmic function, and the inverse
of a logarithmic function is an exponential function.
• The domain of the logarithm function with base b is 0, ∞
• The range of the logarithm function with base b is ( − ∞, ∞)

Examples: a. Evaluate log2 32.

The expression log2 32 asks the question, “To what power the base (2) needs to be raised in
order to get 32 (or 2 raised to what power will be equal to 32)?”. In symbols, we have
logarithmic form exponential form log2 32 = 2 = 32

log2 32 = 25 = 32
Therefore, log2 32 = 5

b. Evaluate log7 49.

The expression log7 49 asks the question, “To what power the base (7) needs to be raised in
order to get 49(or 7 raised to what power will be equal to 49)?”. In symbols, we have
logarithmic form exponential form log7 49 = 7 = 49

log7 49 = 72 = 49
Therefore, log7 49 = 2.

***Try this out! Evaluate: 𝐥 𝟐 .

 CONVERTING LOGARITHMIC EQUATIONS TO EXPONENTIAL EQUATIONS
AND VICE VERSA

When converting equations in logarithmic form to exponential form, you always consider the
given format (or how the values are arranged):
exponential form logarithmic form

B-E-S-A = 𝐥 = B-A-S-E
Base Exponent
Exponent Set everything to be equal to
Set everything to be equal to Answer becomes Argument
Answer Base

Examples. a. Convert the following logarithmic equations to exponential equations.

EXAMPLES
Logarithmic Equation Exponential Equation
Logarithmic Equations Exponential Equations
a.
base 5 raised to log2 =− ANS: 2− =
exponent 3
2 2
log8 4 = ANS: 83 = 4
3
𝐥 𝟐 = =𝟐
*log 1000 = 3 ANS: 103 = 1000
equals an
answer of 125 **ln 1 = 0 ANS: 0
=1

*Common logarithms have base 10. That is, log = log10 .

However, the subscript 10 is omitted when writing common logarithms.
**Natural logarithms have base e. That is, ln = log . “ln” can be read as natural log or “el en”.

b. Convert the following exponential equations to logarithmic equations.

EXAMPLES
Logarithmic Equation Exponential Equation
Logarithmic Equations Exponential Equations
b.
ANS: log3 ( + 8) =− 3 3−3 = ( + 8)
log base 3 of the
argument 81

ANS: log25 (2 − 9) =
𝐥= 25 = (2 − 9)

equals exponent 4 ANS: log4 64 = 2 42 = 64

= ANS: log =− 5 10−5 =

***Try this out! Identify whether the given equation is in
logarithmic form or in exponential form. Then, convert the
given into another form (logarithmic or exponential)
a. log5 (2 − 7 + 1) = 2
b. 106 = 1000000
 LAwS OF LOGARITHMS
Since logarithms are exponents, laws of exponents give rise to laws of logarithms. These
properties can be used to simplify logarithmic expressions and solve logarithmic equations.
Exponents Logarithms
Law/Property/Rule
Symbols Symbols Example

𝐥 () =𝐥 log5 [ + 1 ]
Product Rule ∙ =+
+𝐥 = 𝐥 + 𝐥( + )

=−
Quotient Rule log4
𝐥 =𝐥 −𝐥
= 𝐥( + ) − 𝐥( − )
log3 ( + 2)5
Power Rule =∙ 𝐥= 𝐥
= 𝐥( + )
𝐥=

Identity Rule 1=
𝒍 = 𝐥=

or (𝐥 = )
𝐥=

Zero Property 0= 1 𝒍 = 𝐥=

or (𝐥 = )
Negative 1 −1
Exponent or = log5 −1 =− 𝐥
𝐥−= 𝐥 =− 𝐥
Inverse Property
Common Base
Property of If = , then log3 ( + 3) = log3 5
If 𝐥 = 𝐥 , Then =
Equality = + =
(One-to-one)

𝐥 𝐥 𝐥
Change of Base = = =
𝐥 𝐥 𝐥
𝐥
Note: If your goal is to find exact value of a specific logarithm which cannot be solved by simply
getting its exponential form, change the base to 10 or e since these logarithms can be calculated
on most calculators.
Example: a. Here, we will use base 10.
base 10 is omitted.
𝐥=𝐥
𝐥=
𝐥=𝐥
1.477121
=
0.602060
=.𝟐
b. Same example but this time, we use base e.
𝐥 𝑛𝑠 𝑜 𝐥
𝐥=
𝐥=𝐥
3.401197 𝐥=𝐥
=
1.386294
=.𝟐
Note: For final answers involving logarithms, we limit our answers to four (4) – six (6) decimal places.
 EXPANDING AND CONDENSING LOGARITHMS
EXPANDING

𝐥 ( ) =𝐥 + 𝐥

CONDENSING

Examples. a. Expand the following expressions using the laws of logarithms.

. 𝐥 () . 𝐥(∙ )
Separate the factors using product rule
= log2 + log2 2 + log2 3 = log6 103 + log6 113 product rule

= 𝐥+ 𝐥 power rule
Apply power rule

=𝐥 + 𝐥 +𝐥

.𝐥 .𝐥

= log9 3 − log9 2 quotient rule = log64 − log65 quotient rule

=𝐥 − 𝐥 power rule = 𝐥− 𝐥 power rule

+ 2−
.𝐥 .𝐥 5]
+ −
Apply quotient and product rule
+ log5 −
= log 2 + log − log 3 + log Apply quotient rule and product rule

* can be changed into a fractional exponent. That is, =log5 5 + 2 − log5 3 + 1 2 − 5

−𝐥
= , then apply power rule
1 =log5 5 + log5 + log5 ( + 2) − [ log53
= 2log + log − log 3 + log 2 +1
= 𝐥 + 𝐥𝐥 +
− Remove the parenthesis symbol
𝐥 𝐥 + 𝐥 + 𝐥( + ) − 𝐥 +
or

=𝐥 +𝐥 −𝐥 − 𝐥

Note: It is very important to place parenthesis to distinguish terms in the numerator/denominator.

Question: Can we expand 𝒍 + ?

Answer: No. There is no way to expand the logarithm of a sum or difference inside the
argument of the logarithm.
b. Use the laws of logarithms to condense or express each as a logarithm of a single expression.
 .𝐥 + + 𝐥 − . 𝐥− 𝐥

= log2 +2 + log2 −3 = log3 12 − log3 3 power rule

2 power rule

=𝐥 [ + − ]
product rule
=𝐥 quotient rule
. 𝐥 +𝐥 + 𝐥 .𝐥 +𝐥 − 𝐥

= log3 23 + log3 5 + log342 = log9 + log9 6 − log9 2

= log3 (23 ∙ 5 ∙ 42) may be

=𝐥
= 𝐥 simplified further
. 𝐥 −𝐥 − 𝐥
. 𝐥 + 𝐥 − ( 𝐥 + 𝐥)
= log253 − log242 − log234 1
4
53 =3 log + log − ( log + 2 log )
= log2 4 234 may be simplified further 1 2
4
=log 3 + log − ( log + log )
𝟐
=𝐥

= ( log 3 − log 2)
3
1
log 2 =
4

1
3 4
log 2 =

=𝐥

Note: When we write logarithms as a single expression, the bases of all the logarithms in the expression
must be the same. If not, then the rules like product rule or quotient rule will not be applicable.

Example: 𝐥 + 𝐥

The given example can no longer be written as a single expression since their bases are not the same
(base 2 and base 3). Product rule is not applicable. If that is the case, that will be the final answer.

***Try this out!

2
a. Expand: log5 3 b. Write as a single expression: 2 log + 3log − 5log − 3log
 SOLVING LOGARITHMIC EQUATIONS
GENERAL REMINDERS:
a. A logarithm is an exponent. So, to evaluate the logarithmic expression log , you need to ask the
question, “To what power must b be raised to obtain x?”
b. Always check your answer by substituting the value of x in the original equation. If variables are
found in the argument of the logarithmic equation, always make sure that the argument is
greater than zero (or simply positive). Otherwise, you will have extraneous root/s. Recall: The
logarithm of 0 and any negative number are both UNDEFINED.
c. Analyze each equation before solving.

CASE 1. When there is a single logarithm on ONE SIDE of the equation, then:
we will transform the equation in exponential form.

CASE 2. When there is a single logarithm on EACH SIDE of the equation, then:
we will use the law, If 𝐥 = 𝐥 , 𝒕𝒍 =

CASE 1. When there is a single logarithm on ONE SIDE of the equation, then:
we will transform the equation in exponential form.

Examples. Find the solution set of each logarithmic equation.

. 𝐥 (𝟐 + ) = . 𝐥 (𝟐 + ) = . 𝐥 (𝟐 − ) =

Divide both sides by 4 log5 Divide both sides by 10 log5 Numerical coefficient of log
(2x + 4) = 2 (2x + 3) = 2 expression is already 1.
Transform into exponential form
Transform into exponential form Transform into exponential form 32 = 25 − 2
52 = 2 + 4 52 = 2+3 9 = 25 − 2
25 = 2 + 4 25 = 2 + 3 2= 25 − 9
25 − 3 = 2
25 − 4 = 2 2= 16
22 2
21 = 2
= 2= 16
2 2
= =± 4
11 =
ANS: =
ANS: =±
ANS: =
. 𝐥( + ) = . 𝐥(+ 𝟐 − ) = .𝐥 + =

Transform into exponential form Transform into exponential form Apply power law of logarithm
24 =+1 71 = 2 +
3−3 4 log2 + 1 =8
16 = + 1 7= 2+
3−3
16 − 1 = 2+ 3 − 3 − 7 = 0 Eliminate the numerical coefficient
15 = of the log expression by dividing
2 + 3 − 10 = 0
both sides by 4
+5 −2 =0 log2 + 1 =2
ANS: = 𝟐 + 5 = 0; − 2 = 0
=− 5; = 2 Transform into exponential form
22 = + 1
ANS: =− ; = 4= +1
−1 + 4 =
3 = ANS:
=
CASE 2. When there is a single logarithm on EACH SIDE of the equation, then:
we will use the law, If 𝐥 = 𝐥 , 𝒕𝒍 =

Examples: Find the solution set of each logarithmic equation.

.𝐥 𝟐+ =𝐥 + .𝐥 𝟐+ =𝐥 + . 𝐥 ( + 𝟐) = 𝐥 ( − )

(One-to-one property; (One-to-one property; (One-to-one property;

drop log2) drop log3) drop log4)

3+5= −9 7+3=5+9 9 + 2 = 5 − 12

3− =9−5 7−5=9−3 2 − 5 =− 12 − 9

2 4 2 6 −3 −21
= = =
2 2 2 2 −3 −3

=2 =3 =7

ANS: = ANS: = ANS: =

.𝐥 + = 𝐥 (𝟐 + ) .𝐥 +𝟐+ .𝐥 + = 𝐥 (+ 𝟐 + )
= 𝐥 (+ )
(One-to-one property; (One-to-one property;
drop log = log) (One-to-one property; drop ln = log)
drop ln = log10)
5x + 11 = − 6 + 7 2+ 4 = x2 + 9 + 13
2+ 2x + 1 = x2 + 9
5x − 6x = 7 − 11 4 = 9 + 13
2+1=9
−x = − 4 x −9 = 13 − 4
2=9−1
=4 −9 = 9
2=8
=− 1
ANS: =
=4
ANS: =−
ANS: =
More Complex Examples:
Here, you need to apply some laws of logarithms first. Then, identify whether the equation is
under CASE 1 or CASE 2.

1. 𝐥 ( + ) − 𝐥 ( − ) = 2. 𝐥 ( − 𝟐) = + 𝐥 ( − )

Simplify using quotient law of logarithm Combine like terms and apply laws of logarithm
+1 log2 (x2 − 6x) − log2 1 − x = 3 x2 − 6x
log2 =3 log2 =3
2− 1−x

(CASE 1)
(CASE 1)
Change into exponential form
Change the expression into exponential form
3= +1 3= 2− 6
2 2
2− 1−
2−
6
+1 8=
8= 1−
2−
Cross multiply
Cross multiply 81− = 2− 6
82− = +1
8 − 8 = 2− 6
16 − 8 = + 1
−8 − = 1 − 16 0 = 2− 6 + 8 − 8 0 = 2
−9 =− 15 +2−8

= Solve using factoring

0= + 4 ( − 2)
+4=0 −2=0
ANS: =
=− =2

Note: If we try to check both values of x in number 2, the value of x which is 2 ( = 2), when
substituted in the equation particularly in the logarithmic expression 2 (1 − ), will make the
argument negative.
That is, 2 1 −
21 − 2 𝒍( − )
In logarithms, the logarithm of a negative number is undefined. Thus, the only solution is =− .
3. 𝐥 − = 4. 𝐥𝟐 + 𝐥( − ) = 𝐥(𝟐 + )
This problem does not Apply product rule first
need to be simplified because there is only log x (x − 1) = log(3 + 12)
one logarithm in the problem (CASE 1)
ln 4 − 1 = 3
(CASE 2)
log 4 − 1 =3
Apply One-to-one property (drop log)

Rewrite the problem in exponential form −1 = 3 + 12

3= 2− = 3 + 12
4−1
3+ 1 = 4 2− − 3 − 12 = 0 2 − 4
3+ 1 4 − 12 = 0
=
4 4 Solve using factoring
3+ 1
( − 6)( + 2) = 0
=
−6=0 +2=0
4
= =− 2
+
=
The only possible value is =
or ≈ .
(see note for example 2 above)
Note: The symbol “ ≈ ”indicates approximated value.
5. 𝐥 − 𝐥 − =𝐥 6. 𝐥() − 𝐥( + 𝟐 − ) =
Simplify first using quotient rule
10
2log(x) − log(x2 + 4x − 1) = 0 2log
ln = ln 2
7− x = log(2 + 4 − 1) log x 2 = log(2 +
4 − 1)
(CASE 2)
Apply One-to-one property (drop ln) (CASE 2)
10 Apply One-to-one property (drop log)
2= 2+ 4−1
=2
2− 2= 4−1
7−
10 = 2(7 − ) 0=4−1
10 = 14 − 2 1=4
2 = 14 − 10
=
2=4
=2 ANS: =
ANS: =
7. 𝐥() + 𝐥( + ) = 𝐥( − ) 8. 𝐥() = 𝐥(𝟐 − ) − 𝐥( + )
Simplify first using product rule ln(x)(x + Simplify the right side using quotient rule
3) = ln(20 − 5x) 15 −
ln(x) = ln
Apply One-to-one property (drop ln) +1
()( + 3) = 20 − 5
2+
Apply One-to-one property (drop ln)
3 = 20 − 5
15 −
2+ 3 + 5 − 20 = 0 =
2+
8 − 20 = 0 +1
( + 10)( − 2) = 0 +1 = 15 −
2+
= 15 −
+ 10 = 0 −2=0
=− 10 = 2+
+ − 15 = 0
2 + 2 − 15 = 0

The only possible value is x = 2 ( + 5)( − 3) = 0

(see note for example 2 above) +5=0 −3=0
=− 5 =

The only possible value is x = 3

(see note for example 2 above)
Question: Does every logarithmic equation have a solution?

Answer: No. Keep in mind that we can only apply the logarithm to a positive number. Always check for
extraneous solutions.
9. 23 = 10 10. 9 = 104+6
For this equation there is no way to easily get both For this equation there is no way to easily get both
sides with the same base. Therefore, we need to take sides with the same base. Therefore, we need to take
the logarithm of both sides. the logarithm of both sides.
log 23 = log 10 log 9 = log 104+6

Apply logarithm property that says, Apply logarithm property that says,
log = r log log23 = 𝐥 log10 = () log10+ = +

Since 𝐥 = , we rewrite the equation as Since log10+ = + , we rewrite the equation as log 9
𝐥= =4+6

To solve for the value of x, we isolate x by dividing the To solve for the value of x, we isolate x by applying
right side and left side of the equation with 𝐥 APE and dividing the right side and left side of the
equation with 6.
3 2 1
= Apply M.P.E.
𝐥 𝐥 log 9 = 4 + 6 Apply A.P.E. 6
log 9− 4
= Apply M.P.E.
6
6
=
X= -0.5076
X =1.1073

APPLICATIONS OF LOGARITHMIC FUNCTIONS

Just as many physical phenomena can be modeled by exponential functions, the same is true
for logarithmic functions. Logarithms are useful in measuring the intensities of earthquakes (the Richter
scale), sound (decibels) and acids and bases (pH). We now present yet a different use of a basic
logarithm function, like password strength.
Password strength. The information entropy , in bits, of a randomly generated password consisting of
characters is given by = log2 where is the number of possible symbols for each character in the
password. In general, the higher the entropy, the stronger the password.
a. If a 7-character case-sensitive16 password is comprised of letters and numbers only, find the
associated information entropy.
b. How many possible symbol options per character is required to produce a 7-character
password with an information entropy of 50 bits?
Solutions:
a. There are 26 letters in the alphabet, 52 if upper- and lower-case letters are counted as
different. There are 10 digits (0 through 9) for a total of N = 62 symbols. Since the password is to
be 7 characters long, L = 7.

Thus, H = 7 𝐥 , Evaluating it using change of base . bits

b. We have L = 7 and H = 50 and we need to find N.
Solving the equation using = log2 50 = 7 log2 .

Divide both side by 7 = log2

Change the logarithmic function 2 = to exponential
form
Evaluate ≈ . , so we would need 142 different symbols to choose from.
pH value. In chemistry, the pH value measure of acidity or alkalinity of a solution. It is defined by
the function pH = –log [H3O+] where [H3O+] is the hydronium
ion concentration in moles per liter.
a. Find the pH of a solution with hydronium ion
concentration of 5.1 x 10-4.
b. Determine the hydronium ion concentration of the
solution with pH value of 7.40.
Solutions:
a. To get the pH value of the solution, simply substitute the given to the equation.
pH = –log [H3O+]
pH = –log [5.1 x 10-4]* may directly solve using calculator, for this solution we will solve it using property
of logarithm
pH = −[ log 5.1 + log 10−4] apply product rule pH =
−[ log 5.1 + ( − 4 log 10)] apply power rule pH =
−[0.7076 − 4] pH = −[ − 3.2925] pH = . 𝟐

b. The ph is given to be 7. 40, let x be equal to hydronium ion concentration, thus the working
equation is 7.40 = –log x. Change the expression to exponential form. -7.40 = log x
10 =− 7.40 to find the value of x, in the calculator press shift 10 x then the value
=. –

GRAPH OF LOGARITHMIC FUNCTIONS

Finding the Domain of Logarithmic Functions
When finding the domain of a logarithmic function, it is important to remember that the domain
consists only of positive real numbers. That is, the value you are applying the logarithmic function to,
also known as the argument of the logarithmic function, must be greater than zero.

Examples: 1. = 𝐥( − )

Solution: This function is defined for any values of x such that the argument, in this case 2 − 5 , is
greater than zero. To find the domain, we set up an inequality and solve for x:
− > Show that the argument is greater than zero.
− + > + Add 5 on both sides.
> Simplify

>
Therefore, the domain of = log3 (2 − 5) in interval
notation is ,∞.

1. =𝐥 −

Solution: This function is defined for any values of x such that the argument, in this case 7 − 3 , is
greater than zero. To find the domain, we set up an inequality and solve for x:
− > Show that the argument is greater than zero.
− − > − Subtract 7 on both sides.
− >− Divide by -3, then change the inequality symbol

<
 Therefore, the domain of = log4 ( + 7) in interval notation is −∞, .

2. = 𝐥(− )

Solution: This function is defined for any values of x such that the argument, in this case 2 + 3 − 10, is
greater than zero. To find the domain, we set up an inequality and solve for x:
− > Show that the argument is greater than zero.
− + > + Add 9 on both sides.
> Get the square root of both sides >±

Therefore, the domain of = log5 (2 − 9) in interval notation is −∞, − ∪ (, ∞).

3. = 𝒍 (+ − )

Solution: This function is defined for any values of x such that the argument, in this case 2 + 3 − 10, is
greater than zero. To find the domain, we set up an inequality and solve for x:
+ − > Show that the argument is greater than zero.
+ − > Get the factors of 2 + 3 − 10
<− ; >

Therefore, the domain of = log4 ( + 7) in interval notation is −∞, − ∪ (, ∞).

Sketching the Graph of a Logarithmic Function

To sketch the graph of = 𝐥 you can use the fact that the graphs of inverse functions are reflections of
each other in the line = .

In the same coordinate plane, sketch the graph of each function.

a. = b. () = 𝐥
Solution:
a. For = , construct a table of values. By plotting these points and connecting them with a
smooth curve, you obtain the graph shown in Figure 1.

x -2 -1 0 1 2 3
= 1 2 4 8 Figure 1

b. Because () = 𝐥 is the inverse

function of = , the graph of is
obtained by plotting the points ( , ) and
connecting them with a smooth curve.
The graph of is a reflection of
the graph of in the line = as Figure 2

shown in Figure 1. Sketching the Graph of a Common

Logarithmic Function

Sketch the graph of the common logarithmic function () = 𝒍 .

Identify the vertical asymptote.

Solution:
Begin by constructing a table of values. Note that some of the values can be obtained
without a calculator by using the Inverse Property of
Logarithms. Others require a calculator. Next, plot the points and connect them with a smooth
curve, as shown in Figure 2. The vertical asymptote is x = 0 (y -axis).
Without calculator With calculator

x 1 10 2 5 8

() = 𝒍 -2 -1 0 1 0.301 0.699 0.903

The nature of the graph in Figure 2 is a typical type of functions of the form = 𝐥 , > . They
have one-intercept and one vertical asymptote. Notice how slowly the graph rises for > 1. The basic
characteristics of logarithmic graphs are summarized in Figure 3.

Graph of = 𝐥, >
=𝐥 • Domain: 0, ∞
• Range: −∞, ∞
• x-intercept: 1, 0
• Increasing
• one-to-one function, therefore has an inverse
• y-axis is vertical axis
Figure 3
• continuous
• reflection of the graph of = about the line
=.

Additional insight:
Did you know that the number of cases in the recent COVID-19 pandemic is in logarithmic
pattern? An article entitled “What's a logarithmic graph and how does it help explain the spread of
COVID-19?” explains how logarithm was used to project cases of the said pandemic.

Excerpt from the article: The coronavirus pandemic is undoubtedly the greatest challenge the world has faced
in over a generation. But is the presentation of data about the outbreak leading us to overlook any cause for
optimism? There is an alternative to the commonly seen linear graph that could help give a more detailed
picture. It’s called a logarithmic graph. The most common form of a line-graph has a linear scale. Along the Y
axis, the numbers progress in a steady, linear form – 1, 2, 3, 4, or 10, 20, 30 and so on. But infectious diseases don’t
spread in an even, linear fashion. On a linear scale graph, the rate of growth keeps going up and up – the line
can become almost vertical and appear to go on forever. That can create the impression measures like social
distancing aren’t working. On a logarithmic scale, numbers on the Y-axis don’t move up in equal increments but
instead each interval increases by a set factor – it’s often 10 but could be a factor of 3 or 350 or 3,500, anything
at all. It all depends on what is deemed to be the most effective way of interpreting the data in question. The
Richter scale is logarithmic – an earthquake that measures 6 is 10- times more destructive than one that measures
5.The logarithmic scale is ideal for measuring rates of change, particularly rates of growth, explains
mathematician, teacher, and author of The Life-Changing Magic of Numbers, Bobby Seagull. It “flattens out the
rate of growth so it becomes easier to see,", he says. "On a logarithmic graph of COVID-19 infections, even though
the overall numbers are still increasing, you can see the point at which the rate of growth starts to level off when
that exponential growth has stopped. "At that point, the logarithmic scale makes it possible to see when public
health measures are starting to have the desired effect.

This is how it appears when plotted on the logarithmic scale.