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Why is the logarithm the inverse of the exponential, and what does that reveal about its graph?

Topic 2.10 Inverses of Exponential Functions: construct the inverse of an exponential function as a logarithmic function, and relate the graph, domain and range of each to the other.

A focused answer to AP Precalculus Topic 2.10, covering how the logarithm is the inverse of the exponential, finding the inverse by swapping variables, and how their graphs reflect over y = x with swapped domain and range.

Generated by Claude Opus 4.88 min answer

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  1. What this topic is asking
  2. The logarithm undoes the exponential
  3. Graphs reflect over y = x
  4. Why the asymptotes swap roles
  5. Try this

What this topic is asking

The College Board (Topic 2.10) wants you to see the logarithm as the inverse of the exponential function. You must construct f1f^{-1} for f(x)=bxf(x) = b^x (it is logbx\log_b x), and relate the graph, domain and range of the exponential to those of its logarithmic inverse, including the reflection over the line y=xy = x.

The logarithm undoes the exponential

Finding the inverse uses the swap-and-solve method from Topic 2.8: write y=bxy = b^x, swap to x=byx = b^y, and "solve for yy" by taking a base-bb logarithm, giving y=logbxy = \log_b x.

Graphs reflect over y = x

Because they are inverses, the graphs of bxb^x and logbx\log_b x are mirror images across the line y=xy = x. Every point (p,q)(p, q) on the exponential corresponds to (q,p)(q, p) on the logarithm. The exponential rises steeply and flattens toward its horizontal asymptote; the logarithm, reflected, rises slowly and falls steeply toward its vertical asymptote.

Why the asymptotes swap roles

The reflection over y=xy = x turns horizontal features into vertical ones. The exponential approaches the horizontal line y=0y = 0 as xx \to -\infty; reflecting that line over y=xy = x gives the vertical line x=0x = 0, which is the logarithm's vertical asymptote as x0+x \to 0^+. This is why a logarithm "blows down" to -\infty near x=0x = 0 rather than near a horizontal line. Tracing the asymptote through the reflection, rather than memorizing it separately, ties the two graphs together and explains the logarithm's shape directly from the exponential's.

A clarifying point is that the inverse of an exponential is a logarithm, not a reciprocal or a reflection like bxb^{-x}. The function 2x2^{-x} is a decay curve (a horizontal reflection), and 12x\frac{1}{2^x} is the same thing, but neither undoes 2x2^x. Only log2x\log_2 x returns the original input, which is the defining property of an inverse and the reason logarithms exist as a tool for solving exponential equations in Topic 2.13.

Try this

Q1. What is the inverse of f(x)=10xf(x) = 10^{x}? [1 point]

  • Cue. f1(x)=log10x=logxf^{-1}(x) = \log_{10} x = \log x, the common logarithm.

Q2. The exponential bxb^x has range y>0y > 0. What is the domain of its inverse logbx\log_b x? [1 point]

  • Cue. x>0x > 0: the range of the function becomes the domain of the inverse.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). What is the inverse of f(x)=3xf(x) = 3^{x}? (A) f1(x)=x3f^{-1}(x) = x^{3} (B) f1(x)=log3xf^{-1}(x) = \log_3 x (C) f1(x)=3xf^{-1}(x) = 3^{-x} (D) f1(x)=13xf^{-1}(x) = \frac{1}{3^{x}}
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The correct answer is (B), f1(x)=log3xf^{-1}(x) = \log_3 x.

The inverse of an exponential is the logarithm with the same base. Swapping xx and yy in y=3xy = 3^x gives x=3yx = 3^y, and solving for yy means y=log3xy = \log_3 x. Choice (C) and (D) are reflections or reciprocals, not the inverse.

AP 2024 (style)3 marksSection II (free response, calculator allowed). Let f(x)=exf(x) = e^{x}. (a) State f1(x)f^{-1}(x). (b) Give the domain and range of ff and of f1f^{-1}, and explain how they are related.
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A 3-point question on the exponential-logarithm inverse pair.

(a) Inverse (1 point): f1(x)=lnxf^{-1}(x) = \ln x, the natural logarithm.
(b) Domain and range (2 points): f(x)=exf(x) = e^x has domain all real numbers and range y>0y > 0. Its inverse lnx\ln x has domain x>0x > 0 and range all real numbers. The domain and range swap between a function and its inverse, which is why the exponential's positive range becomes the logarithm's positive domain.

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