Graphs of aˣ, eˣ, lg x and ln x

The problem

Fold a sheet of paper in half and it is 2 layers thick. Fold again: 4, then 8, then 16. Unfold one step at a time and you go back: 8, 4, 2, 1. Doubling and undoing the doubling are the same story told in two directions.

The exponential graph y = aˣ tells the folding story. The logarithm graph tells it backwards, so it is the same curve reflected in the line y = x.

For a > 1: aˣ is always positive, increases as x increases, equals 1 at x = 0, and tends to 0 as x → −∞. Since y = lg x means x = 10ʸ, lg x exists only for x > 0.

Key terms

y = aˣ (a > 1)

Always above the x-axis, passes through (0, 1), and rises faster and faster to the right.

y = eˣ

The same shape as aˣ with a = e ≈ 2.718, so it lies between 2ˣ and 3ˣ.

y = lg x

The mirror of 10ˣ in y = x. It passes through (1, 0) and (10, 1), and is defined only for x > 0.

y = ln x

The mirror of eˣ in y = x. Same shape as lg x, through (1, 0) and (e, 1).

Short questions with model answers

1. Q1. Make a table for y = 2ˣ, x = −4 to 4, and read off its properties.

   • 2⁻⁴ = 1/16 = 0.0625, 2⁻² = 0.25, 2⁰ = 1, 2² = 4, 2⁴ = 16
   • aˣ > 0; increasing; a⁰ = 1; aˣ → 0 as x → −∞

   0.0625, 0.125, 0.25, 0.5, 1, 2, 4, 8, 16

2. Q2. Make a table for y = eˣ, x = −3 to 3, to two decimal places.

   • e⁻³ = 0.05, e⁻² = 0.14, e⁻¹ = 0.37, e⁰ = 1
   • e¹ = 2.72, e² = 7.39, e³ = 20.09

   0.05, 0.14, 0.37, 1, 2.72, 7.39, 20.09

3. Q3. Why does lg x exist only for x > 0? Give lg x at x = 0.1, 1, 2 and 10.

   • y = lg x ⇔ x = 10ʸ > 0
   • lg 0.1 = −1, lg 1 = 0, lg 2 = 0.30, lg 10 = 1

   −1, 0, 0.30, 1

Common mistakes

• ✗ Plotting lg 0 = 0, or drawing the log curve to the left of the y-axis.

  ✓ lg x is undefined at x = 0 and for negative x. As x → 0 from the right, lg x → −∞, so the curve runs down along the y-axis without touching it.

• ✗ Plotting 2⁻³ as −8, below the x-axis.

  ✓ A negative power is a reciprocal: 2⁻³ = 1/8 = 0.125. The graph of aˣ never goes below the x-axis.

• ✗ Copying e³ = 20.07 into your table.

  ✓ e³ = 20.0855, which is 20.09 to two decimal places. Check any table value with a calculator before you plot it.

MCQs

1. 1. On the graph of y = 2ˣ, the y-value at x = −4 is:

   1. (a) −16
   2. (b) −8
   3. (c) 0.0625
   4. (d) 0
   Show answer

   (c) 2⁻⁴ = 1/2⁴ = 1/16 = 0.0625.

2. 2. The domain of lg x is:

   1. (a) all real x
   2. (b) x ≥ 0
   3. (c) x > 0
   4. (d) x > 1
   Show answer

   (c) x = 10ʸ is always positive, so x > 0; lg 0 is undefined.

3. 3. To two decimal places, e³ equals:

   1. (a) 20.07
   2. (b) 20.09
   3. (c) 8.15
   4. (d) 27.00
   Show answer

   (b) 2.71828³ = 20.0855, which rounds to 20.09.

4. 4. For every a > 1, the graph of y = aˣ passes through:

   1. (a) (0, 0)
   2. (b) (1, 0)
   3. (c) (0, 1)
   4. (d) (1, 1)
   Show answer

   (c) a⁰ = 1 for every a, so the point (0, 1) is always on the graph.

5. 5. The graph of y = ln x crosses the x-axis at:

   1. (a) x = 0
   2. (b) x = 1
   3. (c) x = e
   4. (d) never
   Show answer

   (b) ln 1 = 0 because e⁰ = 1. It is the mirror of eˣ's point (0, 1).

Quick revision

• For a > 1, y = aˣ is positive, increasing, passes through (0, 1), and hugs the x-axis as x → −∞.
• lg x and ln x are the mirror images of 10ˣ and eˣ in y = x: defined only for x > 0 and passing through (1, 0).
• Recompute table values before plotting: e³ = 20.09 and e⁻¹ = 0.37.