Mr. Integral
Math notes/Pre-Calculus/Logarithms

The Inverse of an Exponential Function — Cheat sheet

CHEAT SHEET   ·   FREE

Choose “Save as PDF” in the print dialog.
On this page

The move

xx is stuck in an exponent, so the logarithm frees it.

y=ax  ⟺  x=log⁡ayy = a^x \iff x = \log_a y

Method

  1. Write y=f(x)y = f(x).
  2. Peel off everything outside the power first.
  3. Take log⁡a\log_a of both sides. It releases the whole exponent.
  4. Undo what is left, then swap the letters.
FunctionInverse
f(x)=2xf(x) = 2^xf−1(x)=log⁡2xf^{-1}(x) = \log_2 x
f(x)=2x+5f(x) = 2^x + 5f−1(x)=log⁡2(x−5)f^{-1}(x) = \log_2(x - 5)
f(x)=3⋅2xf(x) = 3 \cdot 2^xf−1(x)=log⁡2x3f^{-1}(x) = \log_2 \dfrac{x}{3}
f(x)=2x−1f(x) = 2^{x - 1}f−1(x)=log⁡2x+1f^{-1}(x) = \log_2 x + 1
f(x)=52xf(x) = 5^{2x}f−1(x)=log⁡5x2f^{-1}(x) = \dfrac{\log_5 x}{2}

The whole exponent comes out

log⁡2 ⁣(23x+1)=3x+1\log_2\!\left(2^{3x + 1}\right) = 3x + 1, not 3x3x. Take the exponent out in one piece, then unpick it.

Domain and range change places

domainrange
f(x)=axf(x) = a^xevery real numbery>0y > 0
f−1(x)=log⁡axf^{-1}(x) = \log_a xx>0x > 0every real number

Watch for

  • Adding a constant to axa^x moves the inverse's restriction: f(x)=2x+5f(x) = 2^x + 5 has f−1f^{-1} defined only for x>5x > 5.
  • The graphs are mirror images in the line y=xy = x.