Last modified on March 23rd, 2024

chapter outline

 

Logarithm

Logarithm, often called ‘logs,’ is the power to which a number must be raised to get the result. It is thus the inverse of the exponent and is written as:

ba = x ⇔ logbx = a

Here,

  • ‘b’ is the base
  • ‘a’ is the exponent
  • ‘x’ is the argument

are the 3 parts of a logarithm.

Thus, the logarithm represents the exponent to which a base is raised to yield a given number.

How to Undo Logarithms

For example, we know 43 = 64

Here, using the logarithm, we can answer how many 4s multiply to get 64. 

Since 4 × 4 × 4 = 64, we multiply three 4s to get 64, which is written in the logarithmic form as log4(64) = 3, read as ‘log base 4 of 64 is 3.’

Thus, 43 = 64 ⇔ log4(64) = 3, where the base is 4, and the exponent or power is 3

Here are some examples of conversions from exponential to logarithmic form and vice-versa.

Exponential FormLogarithmic Form
54 = 625log5(625) = 4
152 = 225log15(225) = 2
83 = 512log8(512) = 3

Find the value of log7(343).

Solution:

As we know, 7 × 7 × 7 = 73 = 343
Thus, log7(343) = 3

Convert 35 = 243 in its logarithmic form.

Solution:

As we know, ba = x ⇒ logbx = a
Here, 35 = 243
⇒ log3(243) = 5, the required logarithmic form.

However, the expression logbx has some restrictions, which are as follows:

Restriction on the Variable

For Base

The base ‘b’ of a logarithm is always a positive real number (b > 0) and does not equal 1 (b ≠ 1).

For negative bases, logarithm leads to complex results. Now, let us assume the base is 1, and the equation is: 

log17 = x ⇒ 1x = 7

Since 1 raised to any power yields 1, 1x = 7 is false. Thus, the base does not equal 1.

For Argument

The argument ‘x’ is always a positive real number (a > 0).

Since a positive number (b > 0) is raised to any power, it yields a positive number (bx > 0).

Thus, bx = a follows that a > 0.

Common Logarithms (log) vs Natural Logarithm (ln)

The base of a logarithm can have many positive values except 1. However, two of them are frequently used.

Common Logarithm

The logarithm whose base is 10 is known as the common logarithm or base-10-logarithm. It is often denoted as log(x) without a subscript. 

For example, log10(10000) = log(10000) = 4 ⇔ 104 = 10000

Natural Logarithm

The logarithm with the base ‘e’ (≈ 2.718…, Euler’s number) is the natural logarithm or base-e-logarithm, denoted by ln(x) or loge(x).

For example, ln(e2) = 2 ⇔ e2 = e × e, ln(9) = c ⇔ ec = 9

Rules

Certain rules (also properties or identities) of logarithms are used to simplify, expand, or condense them. 

RulesMathematical Form
Product Rulelogb(xy) = logbx + logby
Quotient Rule${\log _{b}\left( \dfrac{x}{y}\right) =\log _{b}x-\log _{b}y}$
Power Rulelogb(xn) = n logbx
Change of Base Rule${\log _{b}x=\dfrac{\log _{c}x}{\log _{c}b}}$
or, logbx ⋅ logcb = logcx
Zero Rulelogb(1) = 0
Identity Rulelogb(b) = 1
Equality Rulelogbx = logby ⇒ x = y
Inverse Rule${b^{\log _{b}x}=x}$${\log _{b}\left( b^{x}\right) =x}$
Reciprocal Rule${\log _{b}\left( \dfrac{1}{x}\right) =-\log _{b}\left( x\right)}$

Simplifying

Let us simplify a logarithmic function log35 using the appropriate rule(s).

Since logb(xy) = logbx + logby and 35 = 5 × 7 (Using Product rule)

Thus, log35 = log(5 × 7) = log 5 + log 7 ≈ 0.699 + 0.845 (rounded to 3 decimal places)

⇒ log35 ≈ 1.544 (rounded to 3 decimal places)

We observe that logarithms also have decimal values like 1.544, which means 101.544… = 35, and is graphically represented as:

However, negative logarithms are formed when the argument is between 0 and 1. In logbx < 0, for 0 < x < 1, ‘b’ is the base, and ‘x’ is the argument.

For example, log(0.0001) = -4 gives a negative value, and its exponential form, 10-4 = 0.0001, gives a decimal.

Expanding

Let us expand the logarithm expression log(5x4y5).

Here, log(5x4y5)

Using the product rule, we get

log(5) + log(x4) + log(y5)

Using the power rule, we get

log(5) + 4 log(x) + 5 log(y)

Thus, log(5x4y5) = log(5) + 4 log(x) + 5 log(y)

Condensing

Considering the above sum of logarithms log(5) + 4 log(x) + 5 log(y) and condensing it into a single logarithm, we get:

log(5) + 4 log(x) + 5 log(y)

Using the power rule, we get

log(5) + log(x4) + log(y5)

Using the product rule, we get

log(5x4y5)

Thus, log(5) + 4 log(x) + 5 log(y) = log(5x4y5)

Evaluate ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$

Solution:

Here, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$
= ${\log \left( \dfrac{10x^{4}}{y^{2}}\right) +2\log \left( y\right)}$
Using the quotient rule, we get
${\log \left( 10x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$
Using the product rule, we get
${\log \left( 10\right) +\log \left( x^{4}\right) -\log \left( y^{2}\right) +2\log \left( y\right)}$
Using the power rule, we get
${\log \left( 10\right) +4\log \left( x\right) -2\log \left( y\right) +2\log \left( y\right)}$
Since logbb = 1, log(10) = 1
Thus, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$ = 1 + 4log(x)

Expand the expression log2(16x8)

Solution:

Here, log2(16x8)
Using the product law of logarithms, we get
log2(16) + log2(x8)
= 4 + log2(x8)
Using the power law of logarithms, we get
4 + 8 log2(x)
Thus, log2(16x8) = 4 + 8 log2(x)

Condense the expression log5(9x2) – log5(x) + log5(8x4)

Solution:

Here, log5(9x2) – log5(x) + log5(8x4)
First, we use the quotient rule by subtracting the logarithms. The expression is: 
${\log _{5}\left( \dfrac{9x^{2}}{x}\right) +\log _{5}\left( 8x^{4}\right)}$
= ${\log _{5}\left( 9x\right) +\log _{5}\left( 8x^{4}\right)}$
Now, we use the product rule by adding the logarithms. The expression becomes
${\log _{5}\left( 9x\cdot 8x^{4}\right)}$
= ${\log _{5}\left( 72x^{5}\right)}$
Thus, log5(9x2) – log5(x) + log5(8x4) = ${\log _{5}\left( 72x^{5}\right)}$

Solve the equation log(x) + log(3) = log(51) and find the value of x.

Solution:

Here, log(x) + log(3) = log(51)
⇒ log(3x) = log(51) (by the product rule)
Now, canceling the log from both sides, we get
3x = 51
⇒ x = 17
Thus, x = 17

Solving Nested Logarithm

A nested logarithm, also called iterated logarithm or repeated logarithm, represents a logarithm within another logarithm and is denoted as log(log(log…(logx))) or logn(x), where ‘n’ is the level of nesting.

Let us solve the nested logarithm log2(log8(64))

First, we solve the inner logarithm log8(64). Since 82 = 64, log8(64) = 2

Now, solving the outer logarithm, we get log2(2) = 1

Thus, log2(log8(64)) is simplified to 1.

Simplify log5(log3(log7(343)))

Solution:

Here, log5(log3(log7(343)))
Since 73 = 343, log7(343) = 3
Solving the inner logarithm, we get
log5(log3(3))
Since logb(b) = 1, log3(3) = 1
Now, solving further logarithms, we get
log5(1)
= 0

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