How To Solve Logarithmic Equations with Log Laws?

By Robert O

In a previous lesson, you learnt how to find logarithmic base. Therefore, we will assume that you already know all the log laws and their applications. In this tutorial, we will be focusing on how to solve logarithmic equations using those laws that are in your arsenals.

Recap: Log laws and properties

    \[\log _{b} 1=0, \text { from the exponential rule of } b^{0}=1\]

    \[\log _{b} b=1, \text { from the exponential rule of } b^{1}=b\]

    \[\log _{b} b^{x}=x . \text { In general } \log _{b} b^{f(x)}=f(x)\]

    \[b^{\log _{b} x}=x . \text { In general } b^{\log _{b} f(x)}=f(x)\]

    \[\text { If } \log _{b} x=\log _{b} y, \text { then } x=y\]

Product rule:

    \[\log _{b}(m . n)=\log _{b}(m)+\log _{b}(n)\]

Quotient rule:

    \[\log _{b}\left(\frac{m}{n}\right)=\log _{b}(m)-\log _{b}(n)\]

Power rule:

    \[\log _{b}\left(m^{p}\right)=p \log _{b}(m)\]

Change of base rule:

    \[\log _{b} x=\frac{\log _{a} x}{\log _{a} b}\]

How do we solve logarithmic equations using log laws?

We are going to use the laws and rules in the list above to solve all the equations in this tutorial. The same rules and steps apply to any other log equations not covered in this tutorial. Let’s start by looking at some of the examples.

Example 1

Solve for x in:

    \[\log _{2}(x+2)+\log _{2}(3)=\log _{2}(27)\]


The first thing to note is that a logarithmic equation will always have logs of the same base. It is not possible to evaluate logs with different bases as the rules we already know may not apply.

To solve our problem, we use the product rule to reduce the equation on the left side of the equation.

Product rule:

    \[\log _{b}(m . n)=\log _{b}(m)+\log _{b}(n)\]

    \[\log _{2}(x+2)+\log _{2}(3)=\log _{2}(3(x+2))=\log _{2}(3 x+6)\]

The original problem becomes:

    \[\log _{2}(3 x+6)=\log _{2}(27)\]

Since the logs on both sides are of the same base, we drop the logs.

    \[\text { If } \log _{b} x=\log _{b} y, \text { then } x=y\]

    \[3 x+6=27\]



The value of x is only a valid solution if it does not force you to evaluate a logarithm of zero or negatives. Logarithms of zeros and values less than one do not exist. So, always perform the checks to validate the value of x. In this case, x = 7 is valid.

Some logarithmic problems result in two solutions. Check both solutions and state the valid one. One value or both values can be valid.

Example 2


    \[\log _{3}(x+2)-\log _{3}(x)=\log _{3}(2 x-1)-\log _{3}(3 x-12)\]


Apply quotient rule to both sides of the equation.

Quotient rule:

    \[\log _{b}\left(\frac{m}{n}\right)=\log _{b}(m)-\log _{b}(n)\]

    \[\log _{3} \frac{(x+2)}{(x)}=\log _{3} \frac{(2 x-1)}{(3 x-12)}\]

Using the logic:

    \[\text { If } \log _{b} x=\log _{b} y, \text { then } x=y\]

    \[\frac{(x+2)}{(x)}=\frac{(2 x-1)}{(3 x-12)}\]

Carrying out cross-multiplication:

    \[x(2 x-1)=(x+2)(3 x-12)\]

    \[2 x^{2}-x=3 x^{2}-12 x+6 x-24\]

    \[0=x^{2}-5 x-24\]

This is a quadratic equation that we solve using the formula.

    \[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}=\frac{5 \pm \sqrt{25+96}}{2}=\frac{5 \pm 11}{2}\]

    \[x=8 \text { or }-3\]

In this case, we have two values for x, and we have to check their validity. After checking, only x = 8 turns to be valid as the logarithm value for x = -3 is undefined.

Example 3


    \[\log _{6}(10 x)-1=\log _{6}(3 x-1)\]


Group terms with logs on one side and find out which log rules applies.

    \[\log _{6}(10 x)-\log _{6}(3 x-1)=1\]

Applying quotient rule:

    \[\log _{6} \frac{10 x}{(3 x-1)}=1\]

Converting to exponential form:

    \[\frac{10 x}{(3 x-1)}=6^{1}\]

Cross-multiply and solve for x:

    \[10 x=18 x-6\]


A quick check shows that the value for x is valid.


Using laws of logarithms makes it easy to solve equations involving logarithms. We say again that the key to mastering the rules is regular practicing. The rules to solve basic math problems will become part of you after solving a few more problems.

About the Author

This lesson was prepared by Robert O. He holds a Bachelor of Engineering (B.Eng.) degree in Electrical and electronics engineering. He is a career teacher and headed the department of languages and assumed various leadership roles. He writes for Full Potential Learning Academy.