Common log (log₁₀) and natural log (ln x): Visual walkthrough

We need almost nothing to start. Just three plain-English ideas, each drawn:

a function is a machine: feed it a number, it spits one out;
an inverse undoes a machine (put the output back in, get the input);
the slope of a curve at a point is the steepness of the straight line that just grazes it there.

Let's go.

Step 1 — What "slope of a curve" even means

WHAT. Before we touch logarithms, we fix the idea of slope at a point. Take any curve. Pick a point on it. Zoom in. The curve looks straighter and straighter until, super close up, it is a straight line. The steepness of that line is the slope — written $\frac{d y}{d x}$ , read "how much $y$ changes for a tiny change in $x$ ."

WHY this tool. We are hunting for the slope of $ln x$ . So we must first agree what slope means, in a way we can compute. The honest definition uses a tiny step $h$ :

$slope \frac{d y}{d x} = lim_{h \to 0} \frac{f ( x + h ) output a hair to the right - f ( x ) output here}{tiny step to the right h}$

Term by term: the top is rise (how much the output climbed), the bottom $h$ is run (how far right we stepped), and $lim_{h \to 0}$ means "shrink the step to nothing so rise-over-run becomes the true steepness right at the point."

PICTURE. The red line is the secant through two nearby points; as $h$ shrinks it snaps onto the green tangent — the true slope.

Figure — Common log (log₁₀) and natural log (ln x)

Step 2 — What $ln x$ is: the mirror of $e^{x}$

WHAT. The natural log is defined as the inverse of the exponential $e^{x}$ . In plain words: $ln x$ asks " $e$ to what power gives $x$ ?"

$ln x = y ⟺ e^{y} = x$

Reading the two-way arrow: on the left, feed $x$ into the log machine, out comes the power $y$ . On the right, feed that same $y$ into the $e$ -machine, out comes $x$ again. Each undoes the other.

WHY this matters. Inverses have a beautiful geometric fact: their graphs are mirror images across the line $y = x$ . Flip $e^{x}$ over that diagonal and you get $ln x$ . We will squeeze the slope of $ln x$ out of the slope of $e^{x}$ using exactly this mirror — that is the whole plan.

PICTURE. Blue is $e^{x}$ , yellow is $ln x$ , the dashed diagonal is the mirror $y = x$ . Notice a point $(a, b)$ on blue becomes $(b, a)$ on yellow.

Step 3 — The one fact about $e^{x}$ we are allowed to use

WHAT. The exponential $e^{x}$ has one magical property, the very reason $e$ was invented: its slope equals its own height.

$\frac{d}{d x} e^{x} = e^{x}$

At every point on the blue curve, the steepness of the tangent equals the $y$ -value there.

WHY this is fair to assume. This is the defining property of $e$ (built in the parent note from continuous compounding, $e = lim_{n \to \infty} (1 + \frac{1}{n})^{n}$ ). We take it as our single ingredient and derive everything else. If you used base $10$ instead, the slope of $1 0^{x}$ would be $1 0^{x}$ times an ugly leftover factor — we will meet that leftover in Step 7.

PICTURE. At three points the green tangent's steepness (its rise-over-run) is drawn equal to the height of the curve there.

Step 4 — The mirror swaps rise and run

WHAT. Here is the geometric heart. Reflecting across $y = x$ swaps the horizontal and vertical directions. A tangent line to $e^{x}$ with steepness $\frac{rise}{run}$ becomes, after the flip, a tangent line to $ln x$ with steepness $\frac{run}{rise}$ — the reciprocal.

$slope of ln x = \frac{1}{slope of e ^{x} at the mirrored point}$

WHY. Slope is $\frac{up}{across}$ . Mirroring turns "up" into "across" and "across" into "up", so the fraction flips upside down. This is not algebra yet — it is a fact you can see.

PICTURE. The same tangent segment, once as a triangle on blue (rise $R$ , run $1$ ) and again reflected on yellow (rise $1$ , run $R$ ). Its slope went from $R$ to $\frac{1}{R}$ .

Step 5 — Put the two facts together

WHAT. Take a point on the yellow log curve at horizontal position $x$ . Its mirror partner on the blue curve sits at height $x$ (that's what mirroring does — swaps coordinates). By Step 3, the blue tangent there has slope equal to its height, which is $x$ . By Step 4, the yellow slope is the reciprocal:

$\frac{d}{d x} ln x = \frac{1}{= x slope of e ^{x} at that point} = \frac{1}{x}$

Term by term: the mirror sends log-input $x$ to an exponential-output of value $x$ ; the exponential's slope there is that same $x$ (Step 3); flip it (Step 4) and you land on $\frac{1}{x}$ .

WHY it's clean only for base $e$ . The reciprocal was perfect because $e^{x}$ 's slope was exactly its height with no extra factor. That "no extra factor" is the definition of $e$ . Any other base leaves a scar — see Step 7.

PICTURE. One matched pair of points: blue point at height $x$ with tangent slope $x$ ; yellow point at input $x$ with tangent slope $\frac{1}{x}$ .

Step 6 — A pure-algebra check (implicit differentiation)

WHAT. The picture is convincing; here is the same argument in symbols so you can verify it. Start from the definition $y = ln x$ , rewrite as $x = e^{y}$ , and differentiate both sides with respect to $x$ .

$x = e^{y} \frac{d}{d x} \frac{d}{d x} x 1 = slope of e^{y} e^{y} \cdot chain rule \frac{d y}{d x}$

WHY each piece. The left side $\frac{d}{d x} x = 1$ (a thing changing at the same rate as itself). On the right we used Step 3's fact ( $e^{y}$ differentiates to $e^{y}$ ) times $\frac{d y}{d x}$ — the chain rule, because $y$ is itself a function of $x$ , so we must account for how fast $y$ moves. Now solve:

$\frac{d y}{d x} = \frac{1}{e ^{y}} = \frac{1}{x} (since e^{y} = x from the start line) .$

The reciprocal $\frac{1}{e ^{y}}$ is the algebra-twin of the mirror-flip in Step 4, and substituting $e^{y} = x$ is the twin of the coordinate-swap in Step 5. Same story, two languages. See Differentiation of ln x and e^x.

Step 7 — The degenerate & edge cases (never skip these)

Case $x \to 0^{+}$ . As $x$ shrinks toward zero from the right, $\frac{1}{x}$ blows up to $+ \infty$ : the yellow curve becomes vertical. That matches the mirror — $e^{x}$ becomes flat (slope $\to 0$ ) far to the left, and a flat line reflects to a vertical one.

Case $x \to \infty$ . As $x$ grows, $\frac{1}{x} \to 0$ : the log curve flattens out. It keeps rising forever but ever more lazily. Mirror-check: $e^{x}$ gets ever steeper going right, and steep reflects to shallow.

Case $x \leq 0$ (forbidden). There is no point on the yellow curve here at all (Step 2's domain), so asking for its slope is meaningless. For integration we patch this with $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ — the absolute value lets the antiderivative live on the negative side too, but $ln x$ itself still refuses $x \leq 0$ .

Case base $\neq = e$ (why $e$ is special). For a general base $b$ , $\frac{d}{d x} lo g_{b} x = \frac{1}{x l n b}$ . The extra $ln b$ is the "scar" — the leftover factor from $b^{x}$ 's slope not equalling its height. Only when $b = e$ is $ln b = ln e = 1$ , wiping the scar clean. That is why calculus adopts $e$ .

$\frac{d}{d x} lo g_{10} x = \frac{1}{x l n 10} \approx \frac{0.4343}{x}, \frac{d}{d x} ln x = \frac{1}{x} .$

PICTURE. The $\frac{1}{x}$ curve for $x > 0$ , with the vertical blow-up near $0$ (red) and the flattening tail (green), plus the $\frac{1}{x l n 10}$ curve sitting lower to show the scar factor.

The one-picture summary

WHAT. Everything on one canvas: blue $e^{x}$ with its "slope = height" tangent, the dashed mirror $y = x$ , yellow $ln x$ as the reflection, and the resulting $\frac{1}{x}$ slope read straight off a matched pair. Follow the arrow from a height- $x$ point on blue to an input- $x$ point on yellow — the slope flips from $x$ to $\frac{1}{x}$ .

Recall Feynman: the whole walkthrough in one breath

I want the steepness of the " $e$ -to-what?" curve, $ln x$ . I know one lucky fact: the curve $e^{x}$ is so special that at every point its steepness equals its own height — that's literally what makes $e$ the number $e$ . Now, $ln x$ is just $e^{x}$ seen in a mirror placed along the line $y = x$ ; the mirror swaps left–right with up–down. Swapping up and down flips a slope upside-down: rise-over-run becomes run-over-rise. So at the input $x$ , whose mirror partner sits at height $x$ on the $e^{x}$ curve where the slope is $x$ , the log's slope must be the flip of that — namely $\frac{1}{x}$ . Near zero it goes vertical (the mirror of $e^{x}$ going flat), far out it goes flat (the mirror of $e^{x}$ going steep), and it never exists for $x \leq 0$ because $e^{anything}$ can't be zero or negative. Swap the base and a stray factor $ln b$ sneaks in; only $e$ keeps the answer spotlessly $\frac{1}{x}$ .

Connections

Differentiation of ln x and e^x — the home of this result and its companions.
Exponential functions e^x and 10^x — the blue curve we mirrored.
Exponential growth and decay models — where $\frac{1}{x}$ and $e^{x}$ do their real work.
Laws of Logarithms (product, quotient, power) — the algebra behind the change-of-base scar.
Solving exponential equations — same mirror trick, used to solve.

Visual walkthrough — Common log (log₁₀) and natural log (ln x)