The laws of logarithms allow us to rewrite logarithmic expressions to form more convenient expressions. There are seven main laws of logarithms. These seven laws are useful for expanding logarithms, condensing logarithms, and solving logarithmic equations. The laws apply to logarithms of any base, but the same base must be used to apply each law.
Here, we will learn about the main laws of logarithms. In addition, we will look at several examples with answers to understand how to apply these laws to solve algebraic problems.
Descriptions of the laws of logarithms
Remember that a logarithm is the power to which a number must be raised to obtain another number. For example, the base 10 logarithm of 100 is 2, since 10 raised to the power of 2 equals 100:
$latex \log(100)=2$
because:
$latex {{10}^2}=100$
The base is the number that is being raised to a power. We can use logarithms with any base. If we wanted to, we could use the two as the base. For example, the logarithm with base two of eight is equal to three since two raised to the power of three is equal to eight:
$latex \log_{2}(8)=3$
given that
$latex {{2}^3}=8$
Now let’s look at the laws of logarithms.
Law 1: Product Law
This law tells us that the logarithm of a product is equal to the sum of the individual logarithms of the factors:
Law 2: Quotient Law
This law tells us that the logarithm of the quotient of two quantities is equal to the logarithm of the numerator minus the logarithm of the denominator:
Law 3: Power Law
This law tells us that the logarithm of an exponential number is equal to the exponent multiplied by the logarithm of the base:
Law 4: Law of Zero
If we have the logarithm of 1 where the base is $latex b> 0$, but $latex b \neq 0$, the result is equal to zero:
Law 5: Law of identity
The logarithm of the argument (inside the parentheses), where the argument is equal to the base is equal to 1. Since the argument is equal to the base, b must be greater than zero, but not can be equal to 1
Law 6: Law of the logarithm of the exponent
The logarithm of an exponential number, where its base is the same as the base of the logarithm is equal to the exponent:
Law 7: Law of the exponent of a logarithm
Raising the logarithm of a number to its base equals the number:
Laws of logarithms – Examples with answers
The following examples apply the laws of logarithms described above. Each example has its respective solution to look at the reasoning used and the laws of logarithms applied.
EXAMPLE 1
Evaluate the following expression: $latex \log_{2}(8)+\log_{2}(4)$.
Solution
We can apply the product rule to obtain:
$latex \log_{2}(8)+\log_{2}(4)=\log_{2}(8\times 4)$
$latex =\log_{2}(32)$
Now, we can write this expression in its exponential form to get the exponent:
$latex 32={{2}^5}$
Therefore, 5 is the correct answer.
EXAMPLE 2
Evaluate the expression: $latex \log_{3}(162)-\log_{3}(2)$.
Solution
Here, we have a subtraction of logarithms, so we can apply the quotient law:
$latex \log_{3}(162)-\log_{3}(2)=\log_{3}(\frac{162}{2})$
$latex =\log_{3}(81)$
Writing the argument in exponential form, we have:
$latex 81={{3}^4}$
Therefore, the answer is 4.
EXAMPLE 3
Find the value of $latex \log_{\sqrt{2}}(64)$.
Solution
We can use the power law twice to rewrite the expression and make it easier to solve the exercise:
$latex \log_{\sqrt{2}}(64)=\log_{\sqrt{2}}{{(2)}^6}$
$latex =6\log_{\sqrt{2}}(2)$
$latex =6\log_{\sqrt{2}}{{(\sqrt{2})}^2}$
$latex =6\times 2\log_{\sqrt{2}}(\sqrt{2})$
$latex =6\times 2(1)$
$latex =12$
EXAMPLE 4
Simplify the following expression $$\log_{5}(500)-2\log_{5}(2)+\log_{4}(32)+\log_{4}(8)$$
Solution
This expression seems a bit complicated at first glance, but we see that we have only logarithms with two different bases, 4 and 5.
Therefore, we can use the product rule to combine logarithms with base 4 and the quotient rule to combine the logarithms with base 5. However, first we have to apply the power law to the logarithm with base 5:
$$\log_{5}(500)-2\log_{5}(2)+\log_{4}(32)+\log_{4}(8)$$
$$=\log_{5}(500)-\log_{5}{{(2)}^2}+\log_{4}(32)+\log_{4}(8)$$
$$=\log_{5}(500)-\log_{5}(4)+\log_{4}(32)+\log_{4}(8)$$
$latex =\log_{5}(\frac{500}{4})+\log_{4}(32\times 8)$
$latex =\log_{5}(125)+\log_{4}(256)$
Now, we can rewrite the arguments as powers and apply the power law of logarithms:
$latex =\log_{5}({{5}^3})+\log_{4}({{4}^4})$
$latex =3\log_{5}(5)+4\log_{4}(4)$
$latex =3(1)+4(1)$
$latex =7$
EXAMPLE 5
Simplify the expression: $latex 2\log_{3}(5)+\log_{3}(40)-3\log_{3}(10)$.
Solution
We can start by using the power law to rewrite the individual logarithms. Then, we use the product law and the quotient law:
$latex 2\log_{3}(5)+\log_{3}(40)-3\log_{3}(10)$
$latex =\log_{3}({{5}^2})+\log_{3}(40)-\log_{3}({{10}^3})$
$latex =\log_{3}(25)+\log_{3}(40)-\log_{3}(1000)$
$latex =\log_{3}(\frac{25\times 40}{1000})$
$latex =\log_{3}(1)$
Finally, we apply the law of zero to solve:
$latex \log_{3}(1)=0$
EXAMPLE 6
Expand the logarithmic expression: $latex \log_{3}(27{{x}^2}{{y}^5})$.
Solution
We have a product of factors inside the parentheses, so we can apply the product law to write each factor separately. We can simplify individual logarithms when possible.
In this case we use the law of identity to simplify:
$$\log_{3}(27{{x}^2}{{y}^5})=\log_{3}(27)+\log_{3}({{x}^2})+\log_{3}({{y}^5})$$
$latex =\log_{3}({{3}^3})+\log_{3}({{x}^2})+\log_{3}({{y}^5})$
$latex =3\log_{3}(3)+2\log_{3}(x)+5\log_{3}(y)$
$latex =3(1)+2\log_{3}(x)+5\log_{3}(y)$
$latex =3+2\log_{3}(x)+5\log_{3}(y)$
Laws of logarithms – Practice problems
Use the laws of logarithms to solve the following problems. Choose an answer and check it to see that you selected the correct one. If you need help, you can look at the solved examples above or the list of the laws of logarithms.
See also
Interested in learning more about logarithms? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
