Logarithms Explained: Definition, Laws and Solving Equations

What a logarithm is

Exponents and logarithms are two ways of describing the same relationship. The statement $2^3 = 8$ says that raising 2 to the power 3 gives 8. The logarithm turns the question around: to what power must 2 be raised to give 8? The answer is 3, written $\log_2 8 = 3$.

The number $b$ is the base, $x$ is the argument, and $y$ is the exponent you are looking for. The conditions matter. The base must be positive and not equal to 1, because powers of 1 are always 1 and powers of negative numbers are not defined for every real exponent. The argument must be positive, because a positive base raised to any real power is positive.

• $\log_{10} 1000 = 3$, because $10^3 = 1000$.
• $\log_5 \frac{1}{25} = -2$, because $5^{-2} = \frac{1}{25}$.
• $\log_9 3 = \frac{1}{2}$, because $9^{1/2} = \sqrt{9} = 3$.
• $\log_7 1 = 0$ for every allowed base, because $b^0 = 1$, and $\log_b b = 1$ because $b^1 = b$.

Two bases have their own notation. The common logarithm $\log x$ uses base 10, which suits measurements in powers of ten. The natural logarithm $\ln x$ uses base $e \approx 2.71828$, which appears naturally in growth, decay and calculus. Some textbooks and programming languages write $\log$ for the natural log, so check the convention in your course.

The three laws of logarithms

Because logarithms are exponents, the exponent rules translate directly into logarithm rules. For positive $M$ and $N$ and any allowed base:

The product law comes from $b^m b^n = b^{m+n}$: multiplying numbers adds their exponents, so it adds their logarithms. The quotient law comes from dividing powers, and the power law from $(b^m)^k = b^{mk}$. Before calculators, this is exactly why logarithm tables were used: they turned difficult multiplications into additions.

• $\log_2(32 \cdot 4) = \log_2 32 + \log_2 4 = 5 + 2 = 7$, and indeed $128 = 2^7$.
• $\log 50 + \log 2 = \log 100 = 2$.
• $\log_3 54 - \log_3 2 = \log_3 27 = 3$.
• $\ln(e^5) = 5\ln e = 5$.

Expanding and condensing expressions

Expanding uses the laws to split one logarithm into simpler pieces; condensing combines several into one. Both are common exam tasks and both are needed when solving equations.

Coefficients must become exponents before you combine. $2\ln x + \ln y$ is $\ln(x^2 y)$, not $\ln(2xy)$.

Change of base

Calculators usually have keys only for $\log$ and $\ln$. Any other base is handled with the change of base formula, which works with any new base $c$:

For example, $\log_2 10 = \frac{\ln 10}{\ln 2} \approx \frac{2.302585}{0.693147} \approx 3.3219$. A sanity check helps: $2^3 = 8$ and $2^4 = 16$, so the answer must lie between 3 and 4. The formula also shows that $\log_b a = \frac{1}{\log_a b}$, so $\log_8 2 = \frac{1}{3}$ because $\log_2 8 = 3$.

Solving exponential equations

When the unknown is in an exponent, isolate the power, then take a logarithm of both sides so the power law can bring the exponent down.

A frequent error in Example 2 is taking the logarithm before dividing by 5. That is not wrong, but it produces $\ln 5 + (x - 1)\ln 2 = \ln 40$, which takes longer to solve. Isolating the power first keeps the work short.

Solving logarithmic equations

When the unknown is inside a logarithm, combine the logarithms into one, then rewrite the equation in exponential form. Because logarithms of zero and negative numbers are undefined, every candidate solution must be checked in the original equation.

The algebra in Example 3 produced a quadratic. If you need a reminder of how to solve one, see how to solve quadratic equations.

The graph of a logarithm

The function $y = \log_b x$ is the inverse of $y = b^x$, so its graph is the reflection of the exponential graph in the line $y = x$. For $b > 1$ it passes through $(1, 0)$ and $(b, 1)$, rises slowly without bound, and approaches the y-axis as a vertical asymptote. Its domain is $(0, \infty)$ and its range is all real numbers, exactly the reverse of the exponential function's domain and range.

Shifts change the domain. For $y = \ln(x - 3)$, the argument must be positive, so the domain is $(3, \infty)$ and the asymptote moves to $x = 3$. Working out such domains is covered in the guide to domain and range, and the relationship with exponentials is explained further in inverse functions. You can plot several bases together with the graphing calculator.

Where logarithms are used

• Compound growth: an amount growing 5 percent a year doubles when $1.05^t = 2$, so $t = \frac{\ln 2}{\ln 1.05} \approx 14.21$ years.
• Chemistry: pH is $-\log[\mathrm{H}^+]$, so a hydrogen ion concentration of $10^{-3}$ moles per litre gives pH 3, and each pH step is a factor of ten.
• Sound: decibel levels are $10\log_{10}\frac{I}{I_0}$, so a sound 100 times as intense is 20 decibels louder.
• Computer science: binary search on $n$ items needs about $\log_2 n$ steps, so a million items need only about 20.

All four share the same feature: logarithms compress quantities that span many powers of ten into a manageable scale, and they turn multiplicative change into additive change.

Common mistakes

• Writing $\log(a + b) = \log a + \log b$. The product law is about products, and there is no rule for the log of a sum.
• Writing $\frac{\log a}{\log b} = \log\frac{a}{b}$. The left side is a change of base, $\log_b a$.
• Confusing $(\log x)^2$ with $\log x^2 = 2\log x$.
• Forgetting to check solutions of log equations, so an answer that makes an argument negative slips through.
• Mixing up $\log$ and $\ln$ on a calculator.

The logarithm calculator evaluates any base and shows the change of base step, which makes it a quick way to check the examples above.

A study plan for logarithms

Logarithms feel abstract mainly because the notation hides a simple question. A good way to build fluency is to practise translating in both directions until it is automatic. Take ten powers you already know, such as $2^5 = 32$ or $10^{-2} = 0.01$, and rewrite each as a logarithm. Then take ten logarithms and rewrite each as a power. After a short session, evaluating $\log_3 81$ or $\log_{10} 0.001$ becomes a quick mental step instead of a puzzle.

Next, practise the laws on numbers before using them on variables. Checking that $\log_2 8 + \log_2 4$ equals $\log_2 32$ by computing both sides makes the product law believable, and it trains you to spot the false versions listed in the next section. Only then move on to expanding and condensing algebraic expressions.

Finally, solve equations in two families: those with the unknown in an exponent and those with the unknown inside a logarithm. The first family is solved by taking a logarithm, the second by rewriting in exponential form. Recognising which family a problem belongs to is half the work, and checking the domain at the end is the other habit that separates correct answers from nearly correct ones.

Frequently asked questions