Rules or Laws of Logarithms

In this lesson, you’ll be presented with the common rules of logarithms, also known as the “log rules”. These seven (7) log rules are useful in expanding logarithms, condensing logarithms, and solving logarithmic equations. In addition, since the inverse of a logarithmic function is an exponential function, I would also recommend that you go over and master the exponent rules. Believe me, they always go hand in hand.

If you’re ever interested as to why the logarithm rules work, check out my lesson on proofs or justifications of logarithm properties.

But if you think you have a good grasp of the concept, you can simply check out the practice problems below to test your knowledge.

Logarithm Rules Practice Problems


Rules of Logarithms

These are the common log rules that are very useful in the study of algebra. The assumption is that the base b is greater than 1, but b cannot equal 1, and M, N, and k can be any real numbers but M and N must be positive real numbers. Rule 1 is called the Product Rule of logarithm which states that the log of base b of the product of M and M is equal to the log of base b of M plus log of base b of N. In symbol, it is log b (MN) = log b M + log b N. Rule 2 is called the Quotient Rule of logarithm which states that the log of base b of the quotient of M and N is equal to log of base b of M minus log of base b of N. In symbol, we can write this as log b (M/N) = log b M - log b N. Rule 3 is called the Power Rule of logarithm which states that the log of base b of M to the power of k is the k multiplied to the log of base b of M. In symbol, log b (M^k) = (k)*(log b M). Rule 4 is called the Logarithm of 1 Rule which states that the logarithm of 1 of base b is always equal to 0. In symbol, log b (1) = 0. Rule 5 simply states that the logarithm of base b of b is 1, that means, log b (b) = 1. Rule 6 states that the log of base b of b raised to the power of k is k. In math form, log b (b^k) = k. The last rule is rule 7 which states that b raised to the power of the logarithm of k of base b is k. In equation, we can write this as b^[log base b of k) = k.

Descriptions of Logarithm Rules

Rule 1: Product Rule

log base b of (MN) = log base b of (M) + log base b of (N)

The logarithm of the product is the sum of the logarithms of the factors.

Rule 2: Quotient Rule

log base b of (M/N) = log base b of (M) - log base b of (N)

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

Rule 3: Power Rule

log base b of (M^k) = (k) times

The logarithm of an exponential number is the exponent times the logarithm of the base.

Rule 4: Zero Rule

log base b of 1 equals zero where b>0

The logarithm of 1 to any base is always equal to zero. As long as b is positive but b \ne 1.

Rule 5: Identity Rule

log base b of b is equal to 1, where b>1

The logarithm of the argument (inside the parenthesis) wherein the argument equals the base is equal to 1.

Rule 6: Inverse Property of Logarithm

log base b of (b^k) equals k, where b>0

The logarithm of an exponential number where its base is the same as the base of the log is equal to the exponent.

Rule 7: Inverse Property of Exponent

b^(log base b of k) = k

Raising the logarithm of a number to its base is equal to the number.


Examples of How to Apply the Log Rules

Example 1: Evaluate the expression below using Log Rules.

{\log _2}8 + {\log _2}4

Express 8 and 4 as exponential numbers with a base of 2. Then, apply Power Rule followed by Identity Rule. After doing so, you add the resulting values to get your final answer.

log base 2 of 8 + log base 2 of 4 = log base 2 of 2^3 + log base 2 of 2^2 = (3)(log base 2 of 2) + (2) (log base 2 of 2) = 3(1) + (2(1) = 3 + 2 = 5. Therefore the final solution of the logarithm of 8 with base 2 added to the logarithm of 4 with base 2 is equal to 5.

So the answer is \color{blue}5.


Example 2: Evaluate the expression below using Log Rules.

{\log _3}162 - {\log _3}2

We can’t express 162 as an exponential number with base 3. It appears that we’re stuck since there are no rules that can be applied in a direct manner.

The Logarithm Rules can be used in reverse, though! Observe that by using the Quotient Rule reversed, the log expression may be written as a single logarithmic number.

log of base 3 of 162 minus log of base 3 of 2 = log of base 3 of (162/2) = log of base 3 of 81 = log of base 3 of (3^4) = (4) (log of base 3 of 3) = (4)(1) = 1. Therefore the final solution of log of 162 with base 3 minus log of 2 with base 3 is equal to 4.

We did it! By applying the rules in reverse, we generated a single log expression that is easily solvable. The final answer here is \color{blue}4.


Example 3: Evaluate the expression below.

log base 5 of 500 minus 2 times log base 5 of 2 plus log base 4 of 32 plus log base 4 of 8

There appear to be many things going on at the same time. First, see if you can simplify each of the logarithmic numbers. If not, start thinking about some of the obvious logarithmic rules that apply.

By observation, we see that there are two bases involved: 5 and 4. We can start this out by combining the terms that have the same base. Let’s simplify them separately.

For log with base 5, apply the Power Rule first followed by Quotient Rule. For log with base 4, apply the Product Rule immediately. Then get the final answer by adding the two values found.

log 5 (500) - 2 log 2 (2) + log 4 (32) + log 4 (8) = log 5 (125) + log 4 (256) = log 5 (5^3) + log 4 (4^4) = 3(1) + 4(1) = 3+4 =7

Yep, the final answer is \color{blue}7.


Example 4: Expand the logarithmic expression below.

{\log _3}\left( {27{x^2}{y^5}} \right)

A product of factors is contained within the parenthesis. Apply the Product Rule to express them as a sum of individual log expressions. Make an effort to simplify numerical expressions into exact values whenever possible. Use Rule 5 (Identity rule) as much as possible because it can help to simplify the process.

here we are going to expand the logarithm. the log of base 3 of the quantity 27 times x^2 times y^5 = 3 plus 2 times the log of base 3 of x plus 5 times the log of base 3 of y.

I must admit that the final answer appears “unfinished.” But we shouldn’t be concerned as long as we know we followed the rules correctly.


Example 5: Expand the logarithmic expression.

log of base 7 of

The approach is to apply the Quotient Rule first as the difference of two log expressions because they are in fractional form. Then utilize the Product Rule to separate the product of factors as the sum of logarithmic expressions.

the logarithm of  is equal to 2 plus 6 times the log of base 7 of m minus 3 times the log of base 7 of k

Example 6: Expand the logarithmic expression.

log of base 2 of the quantity 12 times w^5 divided by the square root of y

This one has a radical expression in the denominator. Remember that the square root symbol is the same as having a power of {1 \over 2}. Express the radical denominator as {y^{{1 \over 2}}}. Just like problem #5, apply the Quotient Rule for logs and then use the Product Rule.

log base 2 of  = 2 + log base of 2 of 3 + 5 times log base of 2 of w - (1/2) of log base 2 of y

Example 7: Expand the logarithmic expression.

log base 3 of {  /  }

A problem like this may cause you to doubt if indeed you arrived at the correct answer because the final answer can still look “unfinished”.  However, as long as you applied the log rules properly in every step, there’s nothing to worry about.

You might notice that we need to apply the Quotient Rule first because the expression is in fractional form.

log of base 3 of the quantity (18 times the square of x+2) divided by the quantity ( cube of x-2 times the square of x+5) is equal to 2 + log base 3 of 2 + 2 times log base of 3 of (x+2) - 3 times log base of 3 of (x-2) - 2 times log base of 3 of (x+5)

You might also be interested in:

Condensing Logarithms

Expanding Logarithms

Logarithm Explained

Solving Logarithmic Equations

Proofs of Logarithm Properties