Inverse Exponential & Log Functions

Every exponential function of the form f(x) = b^x (where b > 0 and b ≠ 1) has an inverse function, and that inverse is the logarithmic function log_b(x). Understanding this inverse relationship is the key to unlocking why logarithms exist in the first place: they answer the question 'to what power must b be raised to produce x?' This deep connection means that exponential and logarithmic functions are mirror images of each other across the line y = x. Because inverses 'undo' each other, the domain and range of an exponential function become the range and domain of its logarithmic inverse. The exponential f(x) = b^x has domain (−∞, ∞) and range (0, ∞), so its inverse log_b(x) has domain (0, ∞) and range (−∞, ∞). This is why you can never take the logarithm of a negative number or zero — those values were never in the range of the original exponential function. Domain restrictions are not arbitrary rules to memorize; they follow directly from the algebra of inverses. When you find the inverse of a function, you sometimes need to restrict the domain of the original to ensure the inverse is also a function (passes the vertical line test). For exponential functions this restriction is automatic since they are already one-to-one, but recognizing the origin of domain restrictions builds the mathematical maturity needed for calculus and beyond.

Key takeaways
  • Definition of an Inverse Function
  • Swapping x and y to Find the Inverse
  • Verifying with Composition
  • Domain and Range Switch
  • Reflection Across y = x
  • Why One-to-One Matters

Definition of an Inverse Function

Two functions f and g are inverses when f(g(x)) = x for all x in the domain of g, and g(f(x)) = x for all x in the domain of f. Notationally, g = f⁻¹. For exponential and log functions: if f(x) = b^x then f⁻¹(x) = log_b(x).

Swapping x and y to Find the Inverse

To find the inverse of y = b^x algebraically, swap x and y to get x = b^y, then solve for y. By the definition of a logarithm, y = log_b(x). This confirms that log_b(x) is precisely the inverse of b^x.

Verifying with Composition

Compose the two functions to verify: b^(log_b(x)) = x for x > 0, and log_b(b^x) = x for all real x. These two identities are the cancellation equations. For example, 10^(log₁₀(100)) = 10^2 = 100 ✓, and log₁₀(10^5) = 5 ✓.

Domain and Range Switch

Because f(x) = b^x has domain ℝ = (−∞, ∞) and range (0, ∞), the inverse f⁻¹(x) = log_b(x) has domain (0, ∞) and range (−∞, ∞). Inputs to log_b(x) must be strictly positive. There is also a vertical asymptote at x = 0 for all logarithmic functions.

Reflection Across y = x

Graphically, f⁻¹ is obtained by reflecting f over the line y = x. Key point pairs swap: if (0, 1) is on y = b^x, then (1, 0) is on y = log_b(x). If (1, b) is on the exponential, then (b, 1) is on the logarithm. Both graphs share their behavior near these anchor points.

Why One-to-One Matters

An inverse function exists only when the original function is one-to-one (injective) — each output value corresponds to exactly one input. Exponential functions b^x are strictly monotone (always increasing if b > 1, always decreasing if 0 < b < 1), so they are automatically one-to-one and always have an inverse without any extra domain restriction.

The Key Equation

blog⁡b(x)=x(x>0)log⁡b(bx)=x(x∈R)b^{\log_b(x)} = x \quad (x > 0) \qquad \log_b(b^x) = x \quad (x \in \mathbb{R})
  • b — Base (dimensionless): A positive real number not equal to 1; the common base shared by the exponential and its logarithmic inverse.
  • x — Argument (dimensionless): For log_b(x), x must be strictly positive (x > 0). For b^x, x can be any real number.
  • \log_b(x) — Logarithm base b of x (dimensionless): The exponent to which b must be raised to obtain x. It is the inverse function of b^x.

Worked Example

Problem. Let f(x) = 3^x. (a) Find f⁻¹(x). (b) State the domain and range of both f and f⁻¹. (c) Verify by composing f and f⁻¹. (d) Evaluate f⁻¹(27) and f⁻¹(1/9).

f−1(x)=log⁡3(x),f−1(27)=3,f−1 ⁣(19)=−2f^{-1}(x) = \log_3(x),\quad f^{-1}(27) = 3,\quad f^{-1}\!\left(\tfrac{1}{9}\right) = -2

Answer. f⁻¹(x) = log₃(x); domain of f⁻¹ is (0,∞); f⁻¹(27) = 3 and f⁻¹(1/9) = −2.

Figure
Diagram
xyy=xy=3ˣy=log₃x(0,1)(1,0)(3,1)(1,3)1Reflection line y = x maps (a,b) ↔ (b,a)
Origins · 1 minJohn Napier and how this was discovered
JN
John Napier
1550–1617 · Scottish

John Napier invented logarithms around 1614, publishing his work in 'Mirifici Logarithmorum Canonis Descriptio.' His motivation was purely practical: to simplify the enormous calculations required in astronomy and navigation by converting multiplication into addition. Napier's original logarithms were not the base-10 or natural logarithms we use today, but his revolutionary idea that exponents could be used as a calculating tool laid the groundwork for the formal inverse relationship between exponential and logarithmic functions. Napier was also known in his time as 'Marvellous Merchiston' and was rumoured by neighbours to be a magician. He owned a black rooster which he claimed could identify dishonest servants — a clever psychological trick rather than actual magic.

  • Exponential Functions — Logarithmic functions are defined as the inverses of exponential functions, so a solid understanding of b^x is essential before studying log_b(x).
  • Properties of Logarithms — The product, quotient, and power rules for logarithms all follow from the inverse relationship with exponentials and the laws of exponents.
  • Inverse Functions — The general theory of inverse functions — domain/range swaps, the horizontal line test, and composition verification — directly explains why log_b(x) is the inverse of b^x.
  • The Natural Logarithm ln(x) — ln(x) = log_e(x) is the most important special case of the logarithm and is the inverse of e^x, appearing throughout calculus and science.