The log undoes the exponential

The exponential $f(x) = b^x$ takes an exponent and produces a power: it sends 3 to $b^3$. Its inverse must take the power back to the exponent, and that is precisely the log question from Topic 2.9: $\log_b$ of a number IS the exponent that produced it. So $f^{-1}(x) = \log_b x$, not by decree but by definition.

The pairs make it concrete for $b = 2$: the exponential holds $(3, 8)$ because $2^3 = 8$; the log holds the reversed pair $(8, 3)$. Every point on one graph appears on the other with coordinates swapped.

Composing a function with its inverse returns the input, and for this pair the two directions read:

1. $\log_b(b^x) = x$ for every real x: exponentiate, then ask for the exponent, and you get your exponent back.
2. $b^{\log_b x} = x$ for every $x > 0$: find the exponent that hits x, then use it, and you hit x.

Note the mismatched fine print. The first identity works for ALL real x, because $b^x$ accepts any exponent. The second requires $x > 0$, because $\log_b x$ only accepts positive inputs; there is no exponent to find for zero or a negative target. These identities are evaluated by recognition, not computation: $\log_3(3^7) = 7$ on sight, and $10^{\log_{10} 6} = 6$ on sight.

Reflecting $y = b^x$ over the line $y = x$ produces $y = \log_b x$, and the reflection drags every feature along:

1. Domain and range swap. The exponential: domain all reals, range $(0, \infty)$. The log: domain $(0, \infty)$, range all reals. This is WHY logs reject zero and negatives: those numbers were never outputs of $b^x$.
2. The asymptote turns. The exponential hugs the horizontal line $y = 0$ as $x \to -\infty$; its mirror image hugs the VERTICAL line $x = 0$ as inputs shrink toward zero. Same asymptote, reflected.
3. Anchor points swap. $(0, 1)$ on the exponential becomes $(1, 0)$ on the log; $(1, b)$ becomes $(b, 1)$.

And the log's growth is the exponential's growth inverted: instead of exploding, $\log_b x$ (for $b > 1$) increases forever but ever more slowly, with no horizontal ceiling. Slow is not bounded.

The costliest confusion in the unit survives contact with graphs, so kill it numerically. Is $\log_2 x$ the same as $\tfrac{1}{2^x}$? Test $x = 8$: $\log_2 8 = 3$, while $\tfrac{1}{2^8} = \tfrac{1}{256}$. Not close. The reciprocal $2^{-x}$ is itself an exponential (a decaying one); it undoes nothing. The inverse is the log, and the only reliable habit is the round-trip check: does composing return the input? $\log_2(2^8) = 8$. Yes. $\tfrac{1}{2^{2^8}}$? Nonsense. Case closed.