A-level Mathematics

6.1 The functions a^x and e^x — Learn

Prerequisites: the laws of indices for whole-number and fractional powers from Chapter 2; reading a graph and its asymptote from Chapter 2; the idea of a function and its range.

By the end of this section you can: sketch the graph of y=axy = a^x for a positive base aa; state its intercept, range and asymptote; describe how the shape changes with the base; and sketch the graph of y=exy = e^x, explaining what is special about the number ee.

1. An exponential function

An exponential function has the variable in the index. The simplest examples are

y=2x,y=3x,y=(12)x.y = 2^x, \qquad y = 3^x, \qquad y = \left(\frac{1}{2}\right)^x.

The base aa must be positive, because a negative base gives no meaning to powers such as a1/2a^{1/2}. The base also cannot be 11, since 1x=11^x = 1 for every xx and the graph would be a flat line.

Every curve y=axy = a^x with a>0a > 0 passes through the point (0,1)(0, 1), because a0=1a^0 = 1. The yy-intercept is therefore always 11. The curve never touches or crosses the xx-axis: the value is positive for every xx. The line y=0y = 0 is a horizontal asymptote. The domain is all real numbers, and the range is y>0y > 0.

The size of the base sets the shape:

Example 1. Sketch y=4xy = 4^x, and state its intercept, asymptote and range.

The curve rises because 4>14 > 1. It passes through (0,1)(0, 1). The asymptote is y=0y = 0, and the range is y>0y > 0. The graph is steeper than y=2xy = 2^x, because the base is larger.

Example 2. Describe how the graph of y=(13)xy = \left(\dfrac{1}{3}\right)^x differs from the graph of y=3xy = 3^x.

The base 13\dfrac{1}{3} lies between 00 and 11, so y=(13)xy = \left(\dfrac{1}{3}\right)^x falls. It is the reflection of y=3xy = 3^x in the yy-axis, because (13)x=3−x\left(\dfrac{1}{3}\right)^x = 3^{-x}. Both curves pass through (0,1)(0, 1) and both have asymptote y=0y = 0.

Example 3. Work out the value of y=2xy = 2^x when x=3x = 3, x=0x = 0 and x=−2x = -2.

23=8,20=1,2−2=14.2^3 = 8, \qquad 2^0 = 1, \qquad 2^{-2} = \frac{1}{4}.

The values rise as xx rises, and they stay positive even when xx is negative.

Example 4. Write y=5xy = 5^x in the form y=ekxy = e^{kx}, giving kk exactly.

Write the base as a power of ee: 5=eln⁡55 = e^{\ln 5}. Then

5x=(eln⁡5)x=exln⁡5,5^x = \left(e^{\ln 5}\right)^x = e^{x\ln 5},

so k=ln⁡5≈1.609k = \ln 5 \approx 1.609. Every exponential function can be written with base ee in this way.

2. The number e

The number ee is a constant, like π\pi. Its value is

e=2.718281828…e = 2.718281828\ldots

The graph of y=exy = e^x lies between y=2xy = 2^x and y=3xy = 3^x, because ee lies between 22 and 33. It passes through (0,1)(0, 1) and rises for every xx, with asymptote y=0y = 0.

The number ee is chosen for a special reason. At the point (0,1)(0, 1) the tangent to y=exy = e^x has gradient exactly 11. More than that, the gradient of exe^x at every point equals its value at that point. No other base has this property. Section 6.2 makes that statement precise, and it is why ee is called the natural base for growth and decay.

3. Try it: the exponential explorer

Predict first. The figure shows y=axy = a^x with a slider for the base aa, and the dashed curve y=exy = e^x for comparison. Before you move the slider, predict the shape when a=0.5a = 0.5, and predict whether the curve rises or falls when a=3a = 3. Write both predictions down.

Test. Set a=2a = 2 and read the value at x=1x = 1. Change aa to 33 and read it again. Then set a=0.5a = 0.5 and read the value at x=1x = 1 and at x=−1x = -1.

Explain. Write two sentences. The first explains why every curve passes through the same point (0,1)(0, 1), whatever the base. The second explains why the curve falls when aa is between 00 and 11.

Watch out. These mistakes are common.

Reading axa^x as a power of xx. In y=x2y = x^2 the variable is the base; in y=2xy = 2^x the variable is the exponent. The second curve grows far faster. For x=10x = 10, x2=100x^2 = 100 but 210=10242^{10} = 1024.

Thinking the curve touches the xx-axis. The value of axa^x becomes very small as xx becomes very negative, but it never reaches zero. The xx-axis is an asymptote, not a crossing point.

Using a negative or zero base. The expression (−2)x(-2)^x has no real value for x=12x = \dfrac{1}{2}, and 0x0^x is 00 for every positive xx. So the base must be positive.

Re-check: for your sketch, confirm the yy-intercept is (0,1)(0, 1), the curve stays above the xx-axis, and the direction matches the size of the base.

You should now be able to

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