# Log Formulas

Before learning log formulas, let us recall what are logs (logarithms). A logarithm is just another way of writing exponents. When we cannot solve a problem using the exponents, then we use logarithms. There are different log formulas that are derived by using the laws of exponents. Let us learn them using a few solved examples.

## What Are Log Formulas?

Before going to learn the log formulas, let us recall a few things. There are two types of logarithms, common logarithm (which is written as "log" and its base is 10 if not mentioned) and natural logarithm (which is written as "ln" and its base is always "e"). The below log formulas are shown for common logarithms. However, they are all applicable for natural logarithms as well. Here are the log formulas.

Here is the derivation of some important log formulas. We use the laws of exponents in the derivation of log formulas.

### Product Rule of logarithms

The product rule of logs is, log\(_b\) (xy) = log\(_b\) x + log\(_b\) y.

**Derivation:**

Let us assume that log\(_b\) x = m and log\(_b\) y = n. Then by the definition of logarithm,

x = b^{m} and y = b^{n}.

Then xy = b^{m} × b^{n} = b^{m + n} (by a law of exponents, a^{m }× a^{n} = a^{m + n})

Converting xy = b^{m + n} into logarithmic form, we get

m + n = log\(_b\) xy

Substituting the values log\(_b\) x = m and log\(_b\) y = n here,

log\(_b\) (xy) = log\(_b\) x + log\(_b\) y

### Quotient Rule of logarithms

The quotient rule of logs is, log\(_b\) (x/y) = log\(_b\) x - log\(_b\) y.

**Derivation:**

Let us assume that log\(_b\) x = m and log\(_b\) y = n. Then by the definition of logarithm,

x = b^{m} and y = b^{n}.

Then x/y = b^{m} / b^{n} = b^{m - n }(by a law of exponents, a^{m }/ a^{n} = a^{m - n})

Converting x/y = b^{m - n} into logarithmic form, we get

m - n = log\(_b\) (x/y)

Substituting the values log\(_b\) x = m and log\(_b\) y = n here,

log\(_b\) (x/y) = log\(_b\) x - log\(_b\) y

### Power Rule of Logarithms

The power rule of logarithms says log\(_b\) a^{x} = x log\(_b\) a.

**Derivation:**

Let log\(_b\) a = m. Then by the definition of logarithm, a = b^{m}.

Raising both sides by x, we get

a^{x} = (b^{m})^{x}

a^{x} = b^{mx} (by a law of exponents, (a^{m})^{n} = a^{mn})

Converting this back into logarithmic form,

log\(_b\) a^{x} = m x

Substitute m = log\(_b\) a here,

log\(_b\) a^{x} = x log\(_b\) a

### Change of Base Rule of Logarithms

The change of base rule of logs says log\(_b\) a = (log\(_c\) a) / (log\(_c\) b).

**Derivation:**

Assume that log\(_b\) a = x, log\(_c\) a = y, and log\(_c\) b = z.

Converting these into exponential forms,

a = b^{x} ... (1)

a = c^{y} ... (2)

b = c^{z} ... (3)

From (1) and (2),

b^{x} = c^{y}

(c^{z})^{x} = c^{y} (from (3))

c^{zx} = c^{y}

Since the bases are same, the powers also should be the same.

zx = y (or) x = y / z.

Substituting the values of x, y, and z here back,

log\(_b\) a = (log\(_c\) a) / (log\(_c\) b).

We can see the applications of the log formulas in the section below.

## Examples Using Log Formulas

**Example 1: **Convert the following from exponential form to logarithmic form using the log formulas. a) 5^{3} = 125 b) 3^{-3} = 1 / 27.

**Solution:**

Using the definition of the logarithm,

b^{x} = a ⇒ log\(_b\) a = x

Using this,

a) 5^{3} = 125 ⇒ log\(_5\) 125 = 3

b) 3^{-3} = 1 / 27 ⇒ log\(_3\) 1/27 = -3

**Answer**: a) log\(_5\) 125 = 3; b) log\(_3\) 1/27 = -3.

**Example 2: **Compress the following expression as a single logarithm by using rules/properties of exponents. 5 log x + log y - 8 log z.

**Solution:**

To find: The compressed form of the given expression as a single logarithm using log formulas.

5 log x + log y - 8 log z

= (5 log x - 8 log z) + log y (Regrouped the terms)

= (log x^{5} - log z^{8}) + log y (∵ a log x = log x^{a})

= log (x^{5}/z^{8}) + log y (∵ log x - log y = log (x/y) )

= log (x^{5}y/z^{8}) (∵ log x + log y = log (xy) )

**Answer**: 5 log x + log y - 8 log z = log (x^{5}y/z^{8}).

**Example 3: **Find the integer value of log\(_3\) (1/9) using log formulas.

**Solution:**

log\(_3\) (1/9) = log\(_3\) 1 - log\(_3\) 9 (∵ log\(_b\) (x / y) = log\(_b\) x - log\(_b\) y)

= 0 - log\(_3\) 3^{2} (∵ log\(_b\) 1 = 0)

= - 2 log\(_3\) 3 (∵ log\(_b\) a^{x} = x log\(_b\) a)

= -2 (1) (∵ log\(_b\) b = 1)

= -2

**Answer: **log\(_3\) (1/9) = -2.

## FAQs on Log Formulas

### What Are Log Formulas?

The log formulas are related to logarithms and are very helpful while solving the problems of logarithms. Some important log formulas are:

- log\(_b\) (xy) = log\(_b\) x + log\(_b\) y
- log\(_b\) (x / y) = log\(_b\) x - log\(_b\) y
- log\(_b\) ax = x log\(_b\) a
- log\(_b\) a = (log\(_c\) a) / (log\(_c\) b)

### How To Derive Log Formulas?

The laws of exponents are used to derive the log formulas. We also use the definition of logarithm while deriving the log formulas. i.e. we convert the logarithmic form into exponential form and vice versa in the derivation. For a detailed derivation of log formulas, you can refer to the "What Are Log Formulas?" section of this page.

### What Are the Applications of Log Formulas?

The problems that cannot be solved using the exponents' properties can be solved using logs. The log formulas are used to either compress a group of logarithms into a single logarithm or vice versa.

### What Is the Use of the Change of Base Formula (One of the Log Formulas)?

Here is an important use of the change of base formula. Usually, the calculators have options to calculate the logarithms of numbers with base 10 and with base "e". To find the logarithms of numbers with other bases than 10 and "e", we use the change of base formula. For example log\(_2\) 3 = (log 3) / (log 2).