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If you feed a secret number into a machine that multiplies it by 5, how do you get your original number back? You simply divide by 5.
An inverse function is an operation that completely undoes another function, restoring the original input so that fβˆ’1(f(x))=xf^{-1}(f(x)) = x and f(fβˆ’1(x))=xf(f^{-1}(x)) = x.
In 1748, Swiss mathematician Leonhard Euler formally demonstrated that exponential functions and logarithmic functions are exact inverses of each other.
What happens when we graph an exponential curve and its logarithmic counterpart on the exact same grid?

The Geometric Mirror

Because inverse functions swap inputs and outputs, every point (x,y)(x, y) on the curve y=axy = a^x reflects across the line of symmetry y=xy = x to become (y,x)(y, x) on y=log⁑a(x)y = \log_a(x).
This reflection swaps their domain (all valid inputs) and range (all possible outputs): while y=exy = e^x accepts all real numbers and outputs only positives (0,∞)(0, \infty), y=ln⁑(x)y = \ln(x) accepts only positives (0,∞)(0, \infty) and outputs all real numbers.
Graphs give us clear intuition, but how do we prove algebraically that applying one function to the other always collapses right back to xx?

Proving the Inverse Identities

A logarithm log⁑a(b)=c\log_a(b) = c is defined as the exact exponent you must place on base aa to produce bb, meaning log⁑a(b)=cβ€…β€ŠβŸΊβ€…β€Šac=b\log_a(b) = c \iff a^c = b for any base a>0a > 0 where aβ‰ 1a \neq 1.
To prove the first identity alog⁑a(x)=xa^{\log_a(x)} = x, let u=log⁑a(x)u = \log_a(x). Converting this to exponential form gives au=xa^u = x, and substituting log⁑a(x)\log_a(x) back in for uu directly yields alog⁑a(x)=xa^{\log_a(x)} = x for all x>0x > 0.