Home Math blog Logarithms: definition, properties and worked examples Logarithms: definition, properties and worked examples Understand what a logarithm is, when it exists, the product, quotient and power rules, change of base and how to solve exponential equations, with worked examples.
Tutorial Equipe Math Mental Published September 11, 2026 Updated September 17, 2026 2 min read Short answer The logarithm of a in base b is the exponent x such that b^x = a, with b > 0, b ≠ 1 and a > 0. The key rules are log(xy) = log x + log y, log(x/y) = log x − log y, log(xⁿ) = n · log x and the change of base log_b a = log a / log b. Logarithms turn exponential equations into linear ones.
In this article First examples Properties Change of base Exponential equations Common mistakes A logarithm answers a single question: what exponent do I need to raise the base to in order to get this number?
log b a = x ⟺ b x = a \log_b a = x \iff b^x = a log b a = x ⟺ b x = a
with the existence conditions b > 0 b > 0 b > 0 , b ≠ 1 b \neq 1 b = 1 and a > 0 a > 0 a > 0 .
First examples
log 2 8 = 3 \log_2 8 = 3 log 2 8 = 3 , because 2 3 = 8 2^3 = 8 2 3 = 8 .
log 10 1000 = 3 \log_{10} 1000 = 3 log 10 1000 = 3 , because 10 3 = 1000 10^3 = 1000 1 0 3 = 1000 .
log 5 1 = 0 \log_5 1 = 0 log 5 1 = 0 , because 5 0 = 1 5^0 = 1 5 0 = 1 .
log 3 1 9 = − 2 \log_3 \tfrac{1}{9} = -2 log 3 9 1 = − 2 , because 3 − 2 = 1 9 3^{-2} = \tfrac{1}{9} 3 − 2 = 9 1 .
Worked example: log 9 27 \log_9 27 log 9 27
Write both numbers as powers of 3: 9 = 3 2 9 = 3^2 9 = 3 2 and 27 = 3 3 27 = 3^3 27 = 3 3 .
Look for x x x with 9 x = 27 9^x = 27 9 x = 27 : ( 3 2 ) x = 3 3 ⇒ 3 2 x = 3 3 (3^2)^x = 3^3 \Rightarrow 3^{2x} = 3^3 ( 3 2 ) x = 3 3 ⇒ 3 2 x = 3 3 .
2 x = 3 2x = 3 2 x = 3 , so log 9 27 = 3 2 \log_9 27 = \dfrac{3}{2} log 9 27 = 2 3 .
Properties
For valid bases and arguments:
log b ( x ⋅ y ) = log b x + log b y log b x y = log b x − log b y log b x n = n ⋅ log b x log b a = log c a log c b \begin{aligned}
\log_b (x \cdot y) &= \log_b x + \log_b y \\
\log_b \frac{x}{y} &= \log_b x - \log_b y \\
\log_b x^n &= n \cdot \log_b x \\
\log_b a &= \frac{\log_c a}{\log_c b}
\end{aligned} log b ( x ⋅ y ) log b y x log b x n log b a = log b x + log b y = log b x − log b y = n ⋅ log b x = log c b log c a
They all come from the rules of exponents: multiplying powers with the same base adds exponents, and a logarithm is exactly an exponent.
Worked example: log 12 \log 12 log 12 given log 2 ≈ 0.301 \log 2 \approx 0.301 log 2 ≈ 0.301 and log 3 ≈ 0.477 \log 3 \approx 0.477 log 3 ≈ 0.477
Factor: 12 = 2 2 ⋅ 3 12 = 2^2 \cdot 3 12 = 2 2 ⋅ 3 .
log 12 = 2 log 2 + log 3 \log 12 = 2 \log 2 + \log 3 log 12 = 2 log 2 + log 3 .
log 12 ≈ 0.602 + 0.477 = 1.079 \log 12 \approx 0.602 + 0.477 = 1.079 log 12 ≈ 0.602 + 0.477 = 1.079 .
Worked example: log 5 \log 5 log 5 with the same data
5 = 10 2 5 = \dfrac{10}{2} 5 = 2 10 .
log 5 = log 10 − log 2 ≈ 1 − 0.301 = 0.699 \log 5 = \log 10 - \log 2 \approx 1 - 0.301 = 0.699 log 5 = log 10 − log 2 ≈ 1 − 0.301 = 0.699 .
Change of base
Calculators usually have only log \log log (base 10) and ln \ln ln (base e e e ). For any other base, divide:
Worked example: log 2 3 \log_2 3 log 2 3
log 2 3 = log 3 log 2 \log_2 3 = \dfrac{\log 3}{\log 2} log 2 3 = log 2 log 3 .
≈ 0.477 0.301 ≈ 1.585 \approx \dfrac{0.477}{0.301} \approx 1.585 ≈ 0.301 0.477 ≈ 1.585 .
Rough check: 2 1 = 2 < 3 < 4 = 2 2 2^1 = 2 < 3 < 4 = 2^2 2 1 = 2 < 3 < 4 = 2 2 , so the value lies between 1 and 2.
Exponential equations
When you cannot match the bases, take the logarithm of both sides and use the power rule.
Worked example: 2 x = 10 2^x = 10 2 x = 10
log 2 x = log 10 \log 2^x = \log 10 log 2 x = log 10 .
x ⋅ log 2 = 1 x \cdot \log 2 = 1 x ⋅ log 2 = 1 .
x = 1 log 2 ≈ 1 0.301 ≈ 3.32 x = \dfrac{1}{\log 2} \approx \dfrac{1}{0.301} \approx 3.32 x = log 2 1 ≈ 0.301 1 ≈ 3.32 .
Worked example: 3 x + 1 = 81 3^{x+1} = 81 3 x + 1 = 81
81 = 3 4 81 = 3^4 81 = 3 4 , so the bases can be matched: x + 1 = 4 x + 1 = 4 x + 1 = 4 .
x = 3 x = 3 x = 3 .
Common mistakes
Writing log ( a + b ) = log a + log b \log(a + b) = \log a + \log b log ( a + b ) = log a + log b .
Ignoring the existence conditions and accepting solutions that make the argument negative.
Confusing log b x n = n log b x \log_b x^n = n \log_b x log b x n = n log b x with ( log b x ) n (\log_b x)^n ( log b x ) n .
Frequently asked questions Why can't the base of a logarithm be 1? Because 1 raised to any power is 1. The equation 1^x = 5 has no solution and 1^x = 1 has infinitely many, so the logarithm would not be well defined.
Is there a logarithm of a negative number? Not in the real numbers. With a positive base, b^x is always positive, so no real exponent produces a negative number or zero.
What does log mean with no base written? In most school math, log x means base 10 and ln x means base e ≈ 2.718. In computer science, log often means base 2, so check the convention.
Is log(a + b) equal to log a + log b? No. The rule is for products: log(a · b) = log a + log b. For example, log(1 + 1) = log 2 ≈ 0.301, but log 1 + log 1 = 0.
Practice with freshly generated questions The Logarithms unit in Math Mental has short lessons on this topic, with instant feedback and step-by-step solutions.
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