The Inverse of a Logarithmic Function — Cheat sheet

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The move

xx is stuck inside a logarithm, so the exponential frees it.

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

Method

  1. Write y=f(x)y = f(x).
  2. Make the logarithm stand alone: clear coefficients and constants first.
  3. Raise the base to both sides. It releases the whole argument.
  4. Undo what is left, then swap the letters.
FunctionInverse
f(x)=log⁡3xf(x) = \log_3 xf−1(x)=3xf^{-1}(x) = 3^x
f(x)=log⁡2x+1f(x) = \log_2 x + 1f−1(x)=2x−1f^{-1}(x) = 2^{x - 1}
f(x)=log⁡5(x−2)f(x) = \log_5(x - 2)f−1(x)=5x+2f^{-1}(x) = 5^x + 2
f(x)=2log⁡3xf(x) = 2\log_3 xf−1(x)=3x/2f^{-1}(x) = 3^{x/2}
f(x)=log⁡4(3x)f(x) = \log_4(3x)f−1(x)=4x3f^{-1}(x) = \dfrac{4^x}{3}

Stand alone first

23log⁡2x≠x2^{3\log_2 x} \ne x. Divide by the 33 before raising the base:

3log⁡2x=y  ⇒  log⁡2x=y3  ⇒  x=2y/33\log_2 x = y \;\Rightarrow\; \log_2 x = \frac{y}{3} \;\Rightarrow\; x = 2^{y/3}

The whole argument comes out

5log⁡5(x−2)=x−25^{\log_5(x - 2)} = x - 2. The bracket arrives intact; the −2-2 comes off afterwards.

Domain and range change places

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

Which way round

xx sits......so use
in an exponentthe logarithm
inside a logarithmthe exponential