Intersection of a logarithm and a line

An intersection of a logarithm and a horizontal line amounts to solving a logarithmic equation. We find where $y = \ln x$ meets the line $y = 1$ .

Going back to the exponential

At the crossing the $y$ values are equal, so $\ln x = 1$ . Raise $e$ to both sides.

e^{\ln x} = e^{1}

Since $e^x$ and $\ln x$ are inverses, the left side collapses back to $x$ , leaving $x = e$ . The intersection is $(e, 1)$ , with $e \approx 2.718$ .

Solving a logarithmic equation is exactly this act of exponentiating. If $\ln x = k$ then $x = e^k$ , which, given that $\ln x$ answers the question of what power of $e$ yields $x$ , is little more than the definition read back to front.

Always exactly one intersection

The function $y = \ln x$ is defined only for $x > 0$ and increases monotonically, so it never repeats a height and can meet a horizontal line at most once. And because its range is all real numbers, it meets the line $y = k$ at exactly one point for every $k$ . Contrast the exponential, whose range is limited to $y > 0$ and which therefore misses any line with $k \leq 0$ . Being inverses, the two functions have their domains and ranges swapped, and that is what shows up here.

Where the crossings sit

Line	$x$ -coordinate of the intersection	Approximate value
$y = 0$	$e^{0}$	$1$
$y = 1$	$e^{1}$	$2.72$
$y = 2$	$e^{2}$	$7.39$
$y = 3$	$e^{3}$	$20.1$

Each unit of height costs a factor of $e$ in $x$ . How slowly $\ln x$ climbs is written plainly in the spacing of these intersections.

Relation to the base

The base makes no difference to the argument. If $\log_{10} x = 1$ then $x = 10$ ; if $\log_2 x = 1$ then $x = 2$ . In other words, the crossing of a logarithm with the line $y = 1$ always sits at $x$ equal to the base itself. The base of $\ln x$ is $e$ , so the intersection had to be $(e, 1)$ all along. The larger the base, the farther right the crossing, and the more gently the curve climbs.

Confirming the symmetry

The point $(e, 1)$ is the reflection across the line $y = x$ of the point $(1, e)$ , where the exponential $y = e^x$ meets the vertical line $x = 1$ . It is a small confirmation that the graphs of inverse functions are mirror images in $y = x$ . The large dot on the graph is the intersection $(e, 1)$ .