Table of Contents

Last modified on March 23rd, 2024

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:

b^{a} = x ⇔ log_{b}x = 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.

For example, we know 4^{3} = 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 log_{4}(64) = 3, read as ‘log base 4 of 64 is 3.’

Thus, 4^{3} = 64 ⇔ log_{4}(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 Form | Logarithmic Form |
---|---|

5^{4} = 625 | log_{5}(625) = 4 |

15^{2} = 225 | log_{15}(225) = 2 |

8^{3} = 512 | log_{8}(512) = 3 |

**Find the value of log _{7}(343).**

Solution:

As we know, 7 × 7 × 7 = 7^{3} = 343

Thus, log_{7}(343) = 3

**Convert 3 ^{5} = 243 in its logarithmic form.**

Solution:

As we know, b^{a} = x ⇒ log_{b}x = a

Here, 3^{5} = 243

⇒ log_{3}(243) = 5, the required logarithmic form.

However, the expression log_{b}x has some restrictions, which are as follows:

**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:

log_{1}7 = x ⇒ 1^{x} = 7

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

**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 (b^{x} > 0).

Thus, b^{x} = a follows that a > 0.

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

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, log_{10}(10000) = log(10000) = 4 ⇔ 10^{4} = 10000

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

For example, ln(e^{2}) = 2 ⇔ e^{2} = e × e, ln(9) = c ⇔ e^{c} = 9

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

Rules | Mathematical Form |
---|---|

Product Rule | log_{b}(xy) = log_{b}x + log_{b}y |

Quotient Rule | ${\log _{b}\left( \dfrac{x}{y}\right) =\log _{b}x-\log _{b}y}$ |

Power Rule | log_{b}(x^{n}) = n log_{b}x |

Change of Base Rule | ${\log _{b}x=\dfrac{\log _{c}x}{\log _{c}b}}$ or, log _{b}x ⋅ log_{c}b = log_{c}x |

Zero Rule | log_{b}(1) = 0 |

Identity Rule | log_{b}(b) = 1 |

Equality Rule | log_{b}x = log_{b}y ⇒ x = y |

Inverse Rule | ${b^{\log _{b}x}=x}$${\log _{b}\left( b^{x}\right) =x}$ |

Reciprocal Rule | ${\log _{b}\left( \dfrac{1}{x}\right) =-\log _{b}\left( x\right)}$ |

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

Since log_{b}(xy) = log_{b}x + log_{b}y 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 10^{1.544…} = 35, and is graphically represented as:

However, negative logarithms are formed when the argument is between 0 and 1. In log_{b}x < 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.

Let us expand the logarithm expression log(5x^{4}y^{5}).

Here, log(5x^{4}y^{5})

Using the product rule, we get

log(5) + log(x^{4}) + log(y^{5})

Using the power rule, we get

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

Thus, log(5x^{4}y^{5}) = log(5) + 4 log(x) + 5 log(y)

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(x^{4}) + log(y^{5})

Using the product rule, we get

log(5x^{4}y^{5})

Thus, log(5) + 4 log(x) + 5 log(y) = log(5x^{4}y^{5})

**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 log_{b}b = 1, log(10) = 1

Thus, ${\log \left( \dfrac{100x^{4}}{10y^{2}}\right) +2\log \left( y\right)}$ = 1 + 4log(x)

**Expand the expression log _{2}(16x^{8})**

Solution:

Here, log_{2}(16x^{8})

Using the product law of logarithms, we get

log_{2}(16) + log_{2}(x^{8})

= 4 + log_{2}(x^{8})

Using the power law of logarithms, we get

4 + 8 log_{2}(x)

Thus, log_{2}(16x^{8}) = 4 + 8 log_{2}(x)

**Condense the expression log _{5}(9x^{2}) – log_{5}(x) + log_{5}(8x^{4})**

Solution:

Here, log_{5}(9x^{2}) – log_{5}(x) + log_{5}(8x^{4})

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, log_{5}(9x^{2}) – log_{5}(x) + log_{5}(8x^{4}) = ${\log _{5}\left( 72x^{5}\right)}$

**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

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 log^{n}(x), where ‘n’ is the level of nesting.

Let us solve the nested logarithm log_{2}(log_{8}(64))

First, we solve the inner logarithm log_{8}(64). Since 8^{2} = 64, log_{8}(64) = 2

Now, solving the outer logarithm, we get log_{2}(2) = 1

Thus, log_{2}(log_{8}(64)) is simplified to 1.

**Simplify log _{5}(log_{3}(log_{7}(343)))**

Solution:

Here, log_{5}(log_{3}(log_{7}(343)))

Since 7^{3} = 343, log_{7}(343) = 3

Solving the inner logarithm, we get

log_{5}(log_{3}(3))

Since log_{b}(b) = 1, log_{3}(3) = 1

Now, solving further logarithms, we get

log_{5}(1)

= 0