y = aˣ and y = logₐ x: inverse functions mirrored in y = x

Exponentials & logs

How are y = aˣ and y = logₐ x related, and what does the base a decide?

Slide the base a and drag a point along y = aˣ: its mirror image in y = x walks along y = logₐ x. See why both graphs pass through a fixed point, when they rise or fall, and why e is the base whose slope at (0, 1) is exactly 1.

Drag the ringed points on the board.

High school
Exponential function
Logarithmic function
Inverse functions
Monotonicity and the base
The number e

a

0 < a < 1

a > 1

2.0

x_0

1.0

a^x = y \iff x = \log_a y

$x_0$: 1.0
$y_0 = a^{x_0}$: 2.000
$\log_a y_0$: 1.000
$\ln a\ (\text{slope at } (0,1))$: 0.693

As x grows, yincreases (a > 1)

Guided tour

1Two curves, one fixed point each
With $a=2$ , $y=2^x$ passes through $(0,1)$ because $a^0=1$ for every base, and $y=\log_2 x$ passes through $(1,0)$ . Point $P=(1,2)$ sits on the first curve; its mirror $Q=(2,1)$ sits on the second.
2The log undoes the power
Move $P$ to $x_0=3$ : $y_0=2^3=8$ , and $\log_2 8$ reads $3$ again. Swapping the coordinates of every point on $y=a^x$ gives $y=\log_a x$ , which is why the two graphs mirror each other in $y=x$ .
3A base below 1 turns both curves around
At $a=0.5$ , $y=0.5^x$ falls: $x_0=-2$ gives $0.5^{-2}=4$ . Its mirror $y=\log_{0.5}x$ falls too. A base above 1 makes both rise, a base between 0 and 1 makes both fall, and $a=1$ is not allowed.
4Why e is special
The green line touches $y=a^x$ at $(0,1)$ and its slope is $\ln a$ : $0.693$ for $a=2$ . At $a=2.7$ it reads $0.993$ , almost exactly 1. The base that makes it exactly 1 is $e\approx2.718$ , so $e^x$ is the exponential that is its own slope.

Explanation

1
Two ways to read one equation
$a^x=y\iff x=\log_a y\qquad(a>0,\ a\ne1,\ y>0)$
The exponential asks "what is $a$ to the power $x$ ?"; the logarithm asks the reverse, "what power of $a$ gives $y$ ?". So $y=\log_a x$ is the inverse function of $y=a^x$ : its domain $(0,+\infty)$ is the other one's range, and the other way round.
2
Why the graphs mirror in y = x
If $(m,n)$ lies on $y=a^x$ , then $n=a^m$ , so $m=\log_a n$ and $(n,m)$ lies on $y=\log_a x$ . Swapping coordinates is reflection in the line $y=x$ , which is exactly the dashed segment from $P$ to $Q$ : it always crosses $y=x$ at a right angle, at its midpoint.
3
What the base decides
- $a>1$ : both functions are increasing; $a^x\to0$ as $x\to-\infty$ and $\log_a x\to-\infty$ as $x\to0^+$ .
- $0<a<1$ : both are decreasing. $y=\left(\tfrac1a\right)^x$ is the reflection of $y=a^x$ in the $y$ -axis.
- Every exponential passes through $(0,1)$ and every logarithm through $(1,0)$ , whatever the base.
To compare $\log_2 3$ with $\log_3 2$ , use these anchors: $\log_2 3>\log_2 2=1$ , while $\log_3 2<\log_3 3=1$ .
4
The natural base e
Take the slope of $y=a^x$ from its definition: $\dfrac{a^{x+h}-a^x}{h}=a^x\cdot\dfrac{a^h-1}{h}$ . The second factor does not depend on $x$ ; as $h\to0$ it is the slope at $(0,1)$ , the number written $\ln a$ . So the slope everywhere is the height times the slope at $(0,1)$ : $(a^x)'=a^x\ln a$ . Choosing $a=e\approx2.71828$ makes the factor 1, so $(e^x)'=e^x$ . That is why calculus, growth and decay models, and $\ln x=\log_e x$ all use this base.

Check yourself

Q01

Which point lies on $y=\log_a x$ for every allowed base $a$ ?

Q02

For $0<a<1$ , which statement is true?

Q03

Which is larger, $\log_2 3$ or $\log_3 2$ ?

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