Graphing Logarithmic Functions — Cheat sheet

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The shape

y=log⁡axy = \log_a x: domain x>0x > 0, range every real yy, vertical asymptote x=0x = 0.

BaseShape
a>1a > 1rises, slowly and forever
0<a<10 < a < 1falls, slowly and forever

The two are mirror images in the xx-axis, because log⁡1/ax=−log⁡ax\log_{1/a} x = -\log_a x.

Three points fix the curve

(1a,−1),\left(\frac{1}{a}, -1\right), (1,0),(1, 0), (a,1)(a, 1)

For y=log⁡2xy = \log_2 x: (12,−1)\left(\dfrac{1}{2}, -1\right), (1,0)(1, 0), (2,1)(2, 1). Plot them, follow the shape, done.

(1,0)(1, 0) is on every logarithm graph, whatever the base.

Reflection in y = x

y=axy = a^xy=log⁡axy = \log_a x
domainevery real xxx>0x > 0
rangey>0y > 0every real yy
asymptotehorizontal, y=0y = 0vertical, x=0x = 0
passes through(0,1)(0, 1)(1,0)(1, 0)

Every row is the row above with the coordinates swapped.

Shifts

y=log⁡a(x−h)+ky = \log_a(x - h) + k
  • asymptote x=hx = h, domain x>hx > h — the inside constant moves the wall
  • +k+k lifts the curve and leaves the wall alone

Method for a shifted graph

Take f(x)=log⁡2(x−3)+1f(x) = \log_2(x - 3) + 1:

  1. Wall where the argument is zero: x=3x = 3, so domain x>3x > 3
  2. Argument =1= 1 at x=4x = 4: f(4)=0+1=1f(4) = 0 + 1 = 1
  3. Argument =2= 2 at x=5x = 5: f(5)=1+1=2f(5) = 1 + 1 = 2
  4. Base above 11, so it rises from the wall through (4,1)(4, 1) and (5,2)(5, 2)

Traps

  • A negative height never means a negative xx. log⁡3x=−2\log_3 x = -2 gives x=19x = \dfrac{1}{9}, not −9-9.
  • Heights are negative exactly on 0<x<10 < x < 1.
  • The curve is unbounded below near the wall, but never crosses it.
  • log⁡2x\log_2 x gains one unit of height each time xx doubles.