Most difficulty with logarithms comes from treating them as a new species of object with its own arbitrary rulebook. They are not. A logarithm is an exponent — \(\log_b y\) is the answer to one question: b raised to what power gives y? Hold onto that sentence and every rule in this article becomes something you can rederive in a line rather than memorize.

The graphs below stay locked in a mirror. Slide the exponent to move the point along \(y = b^x\); its reflection across the line \(y = x\) rides the logarithm’s graph, because the two curves are the same relationship read in opposite directions.

The solid curve is y = b^x; the lighter curve is its inverse, the logarithm base b; the dashed diagonal is y = x. The two marked points are reflections of each other: if b^a = y, then the logarithm of y is a — one fact, plotted twice, as the panels on either side of the graph spell out. Slide the base through e ≈ 2.718; nothing dramatic marks it, and that is rather the point. Push the base higher and the exponential steepens while the logarithm flattens.

The same fact, written twice

Every logarithmic statement is an exponential statement wearing different clothes:

\[b^a = y \quad\Longleftrightarrow\quad \log_b y = a.\]

So \(2^5 = 32\) and \(\log_2 32 = 5\) are not two facts; they are one fact. When a logarithm looks opaque, convert it. Asked for \(\log_5 125\), do not reach for a rule — ask the question the notation asks: 5 to what power is 125? Three. Asked to solve \(\log_x 49 = 2\), rewrite it as \(x^2 = 49\) and it stops being a logarithm problem at all.

The graphs say the same thing geometrically. Reflecting a point across the line \(y = x\) swaps its coordinates, and swapping coordinates is exactly what the equivalence above does: the point \((a, y)\) sits on the exponential precisely when \((y, a)\) sits on the logarithm. That is what it means for two functions to be inverses, and it is why the logarithm’s graph inherits everything about the exponential’s, sideways: the exponential’s horizontal asymptote becomes the logarithm’s vertical one, and the exponential’s refusal to output negative numbers becomes the logarithm’s refusal to accept them.

The laws are exponent laws, read backward

Exponents turn multiplication into addition: \(b^m \cdot b^n = b^{m+n}\). Since logarithms are exponents, they inherit this directly. If \(m = \log_b x\) and \(n = \log_b y\), then \(xy = b^m b^n = b^{m+n}\), and reading off the exponent of the product gives

\[\log_b(xy) = \log_b x + \log_b y.\]

The other two laws arrive the same way, each from its exponent-law parent: \(\log_b\!\left(\tfrac{x}{y}\right) = \log_b x - \log_b y\) comes from \(b^m / b^n = b^{m-n}\), and \(\log_b(x^k) = k \log_b x\) comes from \((b^m)^k = b^{mk}\). Three laws, zero new memorization — each is a sentence about exponents you already believe.

The errors worth naming are the laws that do not exist. \(\log_b(x + y)\) is not \(\log_b x + \log_b y\) — logarithms convert multiplication to addition, and they do nothing tidy to sums. And \((\log_b x)^2\) is not \(\log_b(x^2)\): the first squares an exponent after the fact, the second doubles it. Writing the parentheses deliberately is not pedantry; it is the whole meaning.

Solving equations: undo in the right order

To solve an exponential equation, take a logarithm; to solve a logarithmic equation, exponentiate. The only technique is deciding what to undo first. Solve

\[5 \cdot 3^t = 40.\]

Isolate the exponential before touching a logarithm: \(3^t = 8\). Now the equation says, in exponential form, exactly what \(t = \log_3 8\) says in logarithmic form — and the change-of-base formula turns that into something a calculator accepts: \(t = \frac{\ln 8}{\ln 3} \approx 1.893\). Check it forward: \(3^{1.893} \approx 8.00\), and \(5 \cdot 8 = 40\).

The order matters. Taking a log of both sides of \(5 \cdot 3^t = 40\) first is legal but produces \(\log 5 + t\log 3 = \log 40\), an extra step that exists only to undo the 5 you could have divided away at the start. Isolate, then invert.

The base e

Among all possible bases, one is special enough to own a key on the calculator: \(e \approx 2.718\), with its logarithm written \(\ln x\). In precalculus, \(e\) earns its place through continuous growth — compounding a rate more and more often walks the growth factor toward \(e\), which is why continuously compounded models are written \(A(t) = A_0 e^{rt}\). Press “set base to e” above and look at the pair of curves; nothing visually dramatic marks the spot, and that is rather the point: \(e\) is not a strange base, just the natural one, and \(\ln\) obeys every law in this article with no special cases.

For those continuing to calculus: that course runs almost entirely on base \(e\), and it will assume the equivalence \(b^a = y \Leftrightarrow \log_b y = a\), the three laws, and clean parenthesizing as reflexes rather than topics. The reflexes are built here.

A self-test at the slider: push the base down toward its lower limit and watch both curves at once. The exponential flattens toward the horizontal line \(y = 1\), and its mirror steepens toward the vertical. At \(b = 1\) exactly, \(b^x\) would be the constant function 1, which fails the horizontal line test and therefore has no inverse. The rule that a logarithm requires \(b \neq 1\) is that collapse, caught one step before it happens.