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169 LearnersLast updated on October 21, 2025

The rules that help to simplify and solve expressions involving logarithms are known as log rules. Log rules, being the inverse of exponent rules, are based on exponent properties. The rules of logarithms are used to expand or combine logarithmic expressions.
Log rules are essential tools for manipulating and simplifying logarithmic expressions. The rules of logarithms are directly derived from the rules of exponents.
Four primary logarithmic rules are commonly applied:
These guidelines are particularly helpful for solving logarithmic equations and simplifying complex logarithmic expressions.
Furthermore, the relationship between exponential and logarithmic forms \((b^x = m ⇔ log_bm=x)\) yields some basic identities:
Working with logarithms in algebra and more complex mathematics is based on these fundamental principles.
In addition to what we have seen already, there are a number of other logarithmic rules. The following table lists every logarithm rule:
Mathematical properties known as the laws of logarithms, or logarithmic rules, make logarithmic expressions and equations easier to understand and solve. Since logarithms are the opposite of exponentiation, these laws are predicated on the exponentiation rules.
Product Law
\({{{{log_{b} {(MN)} = {log_{b} M }+ {log_{b} N}}}}}\)
According to the first law, taking the logarithm after multiplying two numbers together is equivalent to adding their logarithms (of the same base).
Quotient Law
\({{{{{{log_{b} ({M \over N})}} = {{log_{b} M}} - {log_{b} N}}}}}\)
According to the second law, dividing two numbers and taking the logarithm of the result is the same as subtracting their logarithms (again, of the same base).
Power Law
\({{{{{{log_{b} ({M^{p})} = {{p \cdot log_{b}}} M}}}}}}\)
When a number is raised to a power, the logarithm is equal to the product of the exponent with the logarithm of the base expression.
Change of Base Law
\({{{{log_{a} b = {{{log_{c} b } \over {{log_{c}} a}}}}}}}\)
Because calculators usually only have buttons for logarithms in base 10 (common logarithms, represented by the symbol log) and base e (natural logarithms, represented by the symbol ln), the change of base log law is especially helpful. However, you may come across logarithms in other bases in a variety of mathematical problems.
Although log rules seem simple individually, they are often challenging in exams. Usually, you will have to use any number of combinations of logarithmic functions to derive an answer.
Although at first this seems overwhelming, as always, the best technique to handle these issues is to break the question into smaller bits. Solving logarithmic problems uses algebraic manipulation and logarithmic properties. Here is a detailed guide broken out step-by-step:
1: Evaluate Logs
To evaluate logs, if the unknown variable is outside the logarithm, you should use the base of the logarithm to find the value of the logarithm itself.
For instance, if you have \({{( log_b(x) = y )}}\), you can express it as \({{{{{( x = b^y )}}}}}\) to find \({{{{( x )}}}}\). This is straightforward when the base and argument are simple, but complex values may require a calculator.
Step 2: Convert to Exponential Form
When the unknown variable is located within the logarithm, rewrite the equation as an exponential expression.
For instance, if you have the equation \({{{{(\log_b(x) = y )}}}}\), you can rewrite it as \({{{{{(x = b^y )}}}}}\). You can find the unknown variable by taking the base of the logarithm and raising it to the power of the other side of the equation.
Step 3: Logarithm Combination
When an equation contains multiple logarithms, attempt to combine them by utilizing the properties of logarithms.
For instance, two logarithms with the same base can be combined by addition or subtraction. As a result, the problem may become simpler. For example, the expression \({{{{{log_b}(x) + log_{b}(y)}}}}\) can be written as \({{{{{{log_{b}(x) + log_{b}(y) = log_{b}(xy)}}}}}}\) using the product rule.
Step 4: Look for extraneous solutions:
Always verify the validity of the solution you have found. Because they lead to the undefined logarithm of a negative number or zero, some solutions might not be legitimate. These solutions are likely to be discarded as they are referred to as extraneous solutions.
When utilizing calculators that normally only support base 10 (common log) or base 𝑒 (natural log), the Change of Base Rule in logarithms enables you to transform a logarithm with one base into an equivalent expression with a different base. The following is the formula:
\({{{log_{a} b} = {{log_c b} \over {log_c a}}}}\)
In this case, the original base is denoted by 𝑎, the argument by 𝑏, and the new base by 𝑐, which can be any positive number other than 1.
For example, you may rewrite \({{log_2 8}}\) using base 10 as follows to evaluate it: \({{{log 8} \over {log 2}} \approx {{{0.9031} \over {0.3010}}} = {{3}}}\). Similarly, \({{log_3 9}}\), utilizing natural logs, will be \({{{{ln \text { 9}} \over {ln \text { 3}}} \approx {{2.1972} \over {1.0986}} = 2}}\). This rule is useful for solving equations, simplifying expressions, and evaluating non-standard bases.
Mastering logarithms is important as it helps students to solve complex problems, simplify calculations, and understand real-life applications. Here we will learn some tips and tricks to master log rules.
Common errors include misuse of rules, using zero or negative values, and incorrect handling of bases or exponents. By avoiding these mistakes, we can successfully solve logarithmic expressions and also build self-confidence over a period of time.
Logarithmic rules are not limited to mathematics, they are used in various real-life applications. In this section, we will learn some application of logarithmic rules.
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log_10 (5 × 2)
\({{log_{10} 10 = 1 }}\)
Break the multiplication into parts, use approximate values, and add them to get the final answer.
\({{Log_{10} (5 \times 2) = log_{10} 5 + log_{10} 2 }}\)
\({{log_{10} 5 \approx {0.6990}, {\text { and }} {log_{10} 2} \approx 0.3010 }}\)
\({{0.6990 + 0.3010 = 1}}\)
The final result is \({{log_{10} 10 = 1}}\)
Solve using the Division Rule: log_2 (16/4)
2
Apply the log rule.
\({{Log_{2} {({16\over 4})} = {log_2 16} – {log_2 4}}}\)
Step 2: Convert them to the power of 2.
\({{16 = 2^4}}\), so \({{log_2 16 = 4}}\)
Then, \({{2^2 = 4}}\), so \({{log_2 4 = 2}}\)
Step 3: Subtract the end results, that is, 2 from 4
\(4 – 2 = 2\)
Simply log_3 (9)^2
4
Use power rule
\({{Log_3 (9)^2 = 2 log_3 9}}\)
Step 2: We will simplify log3 9:
\({{9 = 3^2}}\), so \({log_3 9 = log_3(3^2) = 2}\)
Step 3: Substitute \({log_3 9 = 2 }\)into the expression
\({2 \cdot log_3 9 = 2 × 2 = 4 }\)
Convert log_4 64 to a common log (base 10).
3
Firstly, we will use the change of
\({ Log_4 64 = {{log_{10} 64} \over {log_{10} 4}}}\)
Step 2: Next we will find the logarithms using a base of 10 to be calculated:
\({Log_{10} 64 \approx 1.8062}\)
\({Log_{10} 4 \approx 0.6021} \)
Step 3: Finally, divide the values.
\({{1.8062 \over 0.6021} \approx 3}\)
The final answer is 3.
If log 3 = 0.477, find the number of digits in 3^(25).
12 digits
Utilize the formula to determine the number of N digits in any number:
Number of digits in N = \({(log_{10} N) + 1}\)
Let’s consider here \({N = 3^{25}} \)
So,
\({Log_{10} (3^{25}) = 25 \cdot log_{10} 3 }\)
\({= 25 \cdot 0.477}\)
\({= 11.925}\)
Now, apply the formula:
Number of Digits = \({⌊11.925⌋ + 1 = 11 + 1 = 12}\)
Therefore, the answer is 12.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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