In this logarithm rules or log rules guide, students and teachers will learn the presented common laws of logarithms, also called ‘log rules’. Mainly, there are four log rules that are helpful in expanding logarithms, condensing logarithms, and solving logarithmic equations. Along with this, you will also find the proofs of these four log rules and additional laws of logarithms for a better understanding of the basic logarithm concept. Whenever you get confused during homework help please check out the** basic logarithm rules or log rules** prevailing here in this article.

## Logarithm Rules Or Log Rules

There are four following math logarithm formulas:

**Product Rule Law:**log_{a}(MN) = log_{a}M + log_{a}N**Power Rule Law:**log_{a }M^{n}= n log_{a}M**Quotient Rule Law:**log_{a}(M/N) = log_{a}M – log_{a}N**Change of Base Rule Law:**log_{a}M = log_{b}M Ã— log_{a}b

**Also Check: Convert Exponentials and Logarithms**

### Descriptions of Logarithm Rules

Here, we have discussed four log rules along with proofs to grasp the concepts easily and become pro in calculating the logarithm problems. Let’s start with proof 1:

#### 1. Logarithm Product Rule:

The logarithm of the multiplication of x and y is the sum of the logarithm of x and the logarithm of y.

log_{b}(x âˆ™ y) = log_{b}(x) + log_{b}(y)

*Proof of Log Product Rule Law:Â *

log_{a}(MN) = log_{a} M + log_{a} N

Let log_{a} M = x â‡’ a sup>x = M

and log_{a} N= y â‡’ ay = N

Now a^{x} âˆ™ a^{y} = MN or, a^{x+y} = MN

Therefore from definition, we have,

log_{a} (MN) = x + y = log_{a} M + log_{a} N [putting the values of x and y]

**Corollary:** The law is true for more than two positive factors i.e.,

log_{a} (MNP) = log_{a} M + log_{a} N + log_{a} P

since, log_{a} (MNP) = log_{a} (MN) + log_{a} P = log_{a} M+ log_{a} N+ log_{a} P

Therefore in general, log_{a} (MNP â€¦…. )= log_{a} M + log_{a} N + log_{a} P + â€¦â€¦.

So, the product logarithm of two or more positive factors to any positive base other than 1 is equal to the sum of the logarithms of the factors to the same base.

**Example:** Calculate log_{10}(8 âˆ™ 4)?

The given expression matches the logarithm product rule.

So apply the log rule and get the result,

log_{10}(8 âˆ™ 4) = log_{10}(8) + log_{10}(4)

#### 2. Logarithm Power Rule:

The logarithm of x raised to the power of y is y times the logarithm of x.

log_{b}(x^{y}) = y âˆ™ log_{b}(x)

**Proof of Log Power Rule Law:**

log_{a}M^{n} = n log_{a} M

Let log_{a} M^{n} = x â‡’ a^{x} = M^{n}

and log_{a} M = y â‡’ a^{y} = M

Now, a^{x} = M^{n} = (a^{y})n = a^{ny}

Therefore, x = ny or, log_{a} M^{n} = n log_{a} M [putting the values of x and y].

**Example:Â **Find log_{10}(2^{9})?

Given log_{10}(2^{9}) is in logarithm power rule. So, apply the log rule and calculate the output:

Hence, log_{10}(2^{9}) = 9âˆ™ log_{10}(2).

#### 3. Logarithm Quotient Rule Formula

The logarithm of the ratio of two numbers is the logarithm of the numerator minus the logarithm of the denominator.

log_{a}(x / y) = log_{a}(x) – log_{a}(y)

**Proof of Log Quotient Rule Formula:**

Let M = a^{x} and N = a^{y}, then it follows that log_{a}(M) = x and log_{a}(N) = y,

We can now prove the quotient rule as follows:

log_{a} (M/N) = log_{a} (a^{x}/a^{y})

= log_{a}(a^{x-y})

= x – y [Put the values of x and y]

= log_{a} M – log_{a} N

**Corollary:** log_{a} [(M Ã— N Ã— P)/(R Ã— S Ã— T)] = log_{a} (M Ã— N Ã— P) – log_{a} (R Ã— S Ã— T)

= log_{a} M + log_{a} N + log_{a} P – (log_{a} R + log_{a} S + log_{a} T)

**Example:Â **Calculate log_{10}(10 / 5)

Now apply the log quotient rule and get the result,

Therefore, log_{10}(10 / 5) = log_{10}(10) – log_{10}(5).

#### 4. Logarithm Base Change Rule:

The logarithm of M for base b is equal to the base a log of M divided by the base a log of b.

log_{b} M = log_{a} M/log_{a} b

*Proof of Change of base Rule Law:*

**log _{a} M = log_{b} M Ã— log_{a} b**

Assume log_{a} M = x â‡’ a^{x} = M,

log_{b} M = y â‡’ b^{y} = M,

and log_{a} b = z â‡’ a^{z} = b.

Now, a^{x}= M = b^{y} – (a^{z})y = a^{yz}

Therefore x = yz or, log_{a} M = log_{b} M Ã— log_{a} b [putting the values of x, y, and z].

**Corollary:**

(i) Putting M = a on both sides of the change of base rule formula [log_{a} M = log_{b} M Ã— log_{a} b] we get,

log_{a} a = log_{b} a Ã— log_{a} b or, log_{b} a Ã— log_{a} b = 1 [since, log_{a} a = 1]

or, **log _{b} a = 1/log_{a} b**

In other words, the logarithm of a positive number a with respect to a positive base b (â‰ 1) is equal to the reciprocal of logarithm of b with respect to the base a.

(ii) On the basis of the log change of base rule formula we get,

**log _{b} M = log_{a} M/log_{a} b**

In other terms, the logarithm of a positive number M in respect of a positive base b (â‰ 1) is equal to the quotient of the logarithm of the number M and the logarithm of the number b both with respect to any positive base a (â‰ 1).

### List of Some Other Logarithm Rules or Log Rules:

If M > 0, N > 0, a > 0, b > 0 and a â‰ 1, b â‰ 1 and n is any real number, then

(i) log_{a} 1 = 0

(ii) log_{a} a = 1

(iii) a ^{loga M} = M

(iv) log_{a} (MN) = log_{a} M + log_{a} N

(v) log_{a} (M/N) = log_{a} M – log_{a} N

(vi) log_{a} M^{n} = n log_{a} M

(vii) log_{a} M = log_{b} M Ã— log_{a} b

(viii) log_{b} a Ã— log_{a} b = 1

(ix) log_{b} a = 1/log_{a} b

(x) log_{b} M = log_{a} M/log_{a} b

### Solved Examples of How to Apply the Log Rules or Logarithm Rules

1. Evaluate the expression ie., log_{2} 4 + log_{2} 8 by using log rules.

**Solution:**

Given expression is log_{2} 4 + log_{2} 8

First, express 4 and 8 as exponential numbers with a base of 2. Next, apply the logarithm power rule formula followed by the identity rule. Once you finished that, add the resulting values to find the final answer.

log_{2} 4 + log_{2} 8

= log_{2} 2^{2} + log_{2}2^{3} [apply power rule]

= 2 log_{2} 2 + 3 log_{2} 2 [apply identity rule]

= 2(1) + 3(1)

= 2+3

= 5

Hence, the answer for the given expression log_{2} 4 + log_{2} 8 is **5.**

2. Evaluate the expression with Log Rules: log_{3} 162 – log_{3} 6

**Solution:**

log_{3} 162 – log_{3} 6

Now, we can’t express the 162 as an exponential number with base 3. Don’t worry, we have another way of solving the expression.

It is possible by applying the log rules in the reverse process. Yes, we can also apply the logarithm rules in reverse if not solved in a direct manner.

Remember that the log expression can be stated as one or a single logarithm number via using the backward Quotient Rule Law. Sounds different right, but so easy to calculate.

Take the given expression, log_{3} 162 – log_{3} 6

= log_{3} (162/6)

= log_{3} (27)

= log_{3} (3^{3})

= 3 log_{3} 3

= 3(1)

= 3

By applying the rules in reverse, we get the result as 3 for the given expression log_{3} 162 – log_{3} 6.

Hence, log_{3} 162 – log_{3} 6 =** 3.**