Exponential Functions and e – A Level Maths Year 1

1 The Graph of $y = a^x$

An exponential function has the variable in the index: $y = a^x$, where the base $a$ is a positive constant and $a \neq 1$. Compare $x^2$ (variable in the base — a power function) with $2^x$ (variable in the index — an exponential function).

DfE content area F asks you to know and use the function $a^x$ and its graph, where $a$ is positive. Every graph of this type has the same four features.

Feature	For $y = a^x$	Why
$y$-intercept	$(0,\,1)$	$a^0 = 1$ for every base
Sign	$y > 0$ for all $x$	A positive number raised to any power is positive
Asymptote	$y = 0$ (the $x$-axis)	$a^x$ gets ever closer to $0$ but never reaches it
Shape	Growth if $a > 1$; decay if $0 < a < 1$	Each step of $1$ in $x$ multiplies $y$ by $a$

$y = a^{-x}$ is $y = \left(\tfrac{1}{a}\right)^x$. Because $a^{-x} = \left(a^{-1}\right)^x$, the decay curve $y = \left(\tfrac12\right)^x$ is the same as $y = 2^{-x}$ — the reflection of $y = 2^x$ in the $y$-axis.

Worked Example 1 — Comparing exponential graphs

On the same axes sketch $y = 3^x$ and $y = 3^{-x}$. State the coordinates of any intersection and the equation of the asymptote.

①$y = 3^x$: through $(0,1)$, rising steeply to the right, approaching the $x$-axis to the left. It passes through $(1, 3)$ and $(-1, \tfrac13)$.

②$y = 3^{-x}$ is its reflection in the $y$-axis: through $(0,1)$, $(-1, 3)$ and $(1, \tfrac13)$.

③They meet where $3^x = 3^{-x}$, i.e. $x = -x$, so $x = 0$: the single point $(0, 1)$.

④Both have asymptote $y = 0$.

2 The Gradient of an Exponential, and the Number e

Exponential graphs have a remarkable property: the gradient at any point is proportional to the $y$-value there. The steeper part of the curve is exactly where the curve is highest.

Measuring the gradient of $y = a^x$ numerically (using a very short chord) gives:

Function	Gradient at $(0, 1)$	Gradient at any $x$
$y = 2^x$	$\approx 0.693$	$\approx 0.693 \times 2^x$
$y = 3^x$	$\approx 1.099$	$\approx 1.099 \times 3^x$
$y = e^x$	exactly $1$	exactly $e^x$

The multiplier is below $1$ for base $2$ and above $1$ for base $3$, so somewhere between $2$ and $3$ there is a base for which the multiplier is exactly $1$. That base is the number \[ e = 2.718\,281\,828\ldots \] It is irrational, like $\pi$. The function $y = e^x$ is called the exponential function, and it is the only exponential whose gradient equals its own value everywhere.

The gradient of e^kx

If $y = e^{kx}$ then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = k\,e^{kx}$ (in particular $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^x\right) = e^x$)

Constants come along for the ride. $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(Ae^{kx}\right) = Ak\,e^{kx}$. So for $y = Ae^{kx}$, the gradient is $k$ times the $y$-value: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky$. This is the reason exponentials model anything whose rate of change is proportional to its size.

Worked Example 2 — Gradient at a point

Find the gradient of $y = e^{3x}$ at $x = 0$ and at $x = \ln 2$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3e^{3x}$

②At $x = 0$: $3e^0 = 3$.

③At $x = \ln 2$: $3e^{3\ln 2} = 3\left(e^{\ln 2}\right)^3 = 3 \times 2^3 = 24$.

$e^{\ln 2} = 2$ because $\ln$ undoes $e$ — see Section 4.

Worked Example 3 — Where is the gradient a given value?

The curve $y = 4e^{-0.5x}$. Find the coordinates of the point where the gradient is $-1$.

①$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4 \times (-0.5)e^{-0.5x} = -2e^{-0.5x}$

②$-2e^{-0.5x} = -1 \;\Rightarrow\; e^{-0.5x} = \tfrac12$

③Take $\ln$: $-0.5x = \ln\tfrac12 = -\ln 2$, so $x = 2\ln 2 \approx 1.386$.

④$y = 4e^{-\ln 2} = 4 \times \tfrac12 = 2$. The point is $(2\ln 2,\; 2)$.

Quick check: gradient $= ky = -0.5 \times 2 = -1$ ✓

Worked Example 4 — A sum of exponentials

Find the gradient of $y = e^{2x} + 3e^{-x}$ at the point where it crosses the $y$-axis.

①Differentiate each term: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2e^{2x} - 3e^{-x}$

②The $y$-axis is $x = 0$, where $y = 1 + 3 = 4$.

③Gradient $= 2(1) - 3(1) = -1$.

3 Sketching Transformed Exponentials

Most exam graphs are of the form $y = A + Be^{kx}$. Three questions pin the sketch down.

Asymptote: the exponential term tends to $0$ at one end (as $x \to -\infty$ if $k > 0$, as $x \to +\infty$ if $k < 0$), so $y \to A$. The asymptote is $y = A$.
$y$-intercept: put $x = 0$: $y = A + B$.
$x$-intercept (if any): solve $A + Be^{kx} = 0$ using $\ln$.

Worked Example 5 — Sketching $y = 5 - 2e^x$

Sketch $y = 5 - 2e^x$, showing the asymptote and the points where it crosses the axes.

①As $x \to -\infty$, $e^x \to 0$ so $y \to 5$. Asymptote $y = 5$.

②$x = 0$: $y = 5 - 2 = 3$, so $(0, 3)$.

③$y = 0$: $2e^x = 5$, $e^x = 2.5$, $x = \ln 2.5 \approx 0.916$.

④As $x$ increases, $-2e^x$ becomes large and negative, so the curve falls steeply.

The curve never touches its asymptote. Draw it getting close to $y = 5$ without crossing or meeting it, and label the asymptote with its equation.

4 $y = \ln x$, the Inverse of $y = e^x$

The natural logarithm $\ln x$ is the logarithm to base $e$: $\ln x = \log_e x$. It undoes the exponential function: \[ y = e^x \iff x = \ln y. \] So its graph is the reflection of $y = e^x$ in the line $y = x$.

	$y = e^x$	$y = \ln x$
Domain (inputs)	all real $x$	$x > 0$ only
Range (outputs)	$y > 0$	all real $y$
Crosses an axis at	$(0, 1)$	$(1, 0)$
Asymptote	$y = 0$	$x = 0$

Inverse facts

$e^{\ln x} = x$ (for $x > 0$) $\ln\left(e^x\right) = x$ (for all $x$) $\ln 1 = 0$ $\ln e = 1$

Worked Example 6 — Using the inverse

(a) Solve $e^x = 7$. (b) Solve $\ln x = 3$. Give exact answers and then decimals.

①(a) Take $\ln$ of both sides: $x = \ln 7 \approx 1.946$.

②(b) Apply $e$ to both sides: $x = e^3 \approx 20.09$.

"Exact" means leave it as $\ln 7$ or $e^3$. Only round if the question asks for a decimal.

Worked Example 7 — Sketching a translated log graph

Sketch $y = \ln(x + 2)$, giving the asymptote and the intercepts.

①This is $y = \ln x$ translated $2$ units left, so the asymptote moves from $x = 0$ to $x = -2$.

②$x$-intercept: $\ln(x+2) = 0 \Rightarrow x + 2 = 1 \Rightarrow x = -1$.

③$y$-intercept: $y = \ln 2 \approx 0.693$.

④The curve exists only for $x > -2$, plunging down the asymptote on the left and rising ever more slowly on the right.

5 Quick Reference

$y = a^x$

Through $(0,1)$; always positive; asymptote $y=0$.

Growth or decay

$a > 1$ grows; $0 < a < 1$ decays.

Reflection

$a^{-x} = (1/a)^x$: reflect in the $y$-axis.

The number e

$e \approx 2.71828$; gradient of $e^x$ is $e^x$.

Gradient rule

$\dfrac{\mathrm{d}}{\mathrm{d}x}\left(Ae^{kx}\right) = Ake^{kx}$

Proportional rate

$y = Ae^{kx}$ gives $\dfrac{\mathrm{d}y}{\mathrm{d}x} = ky$.

Natural log

$\ln x = \log_e x$; inverse of $e^x$.

$y = \ln x$

Only $x > 0$; through $(1,0)$; asymptote $x = 0$.

Undoing

$e^{\ln x} = x$ and $\ln e^x = x$.

$y = A + Be^{kx}$

Asymptote $y = A$; $y$-intercept $A + B$.

6 Practice Questions

Question 1

Sketch $y = 4^x$. State the coordinates of the $y$-intercept and the equation of the asymptote, and find $y$ when $x = -\tfrac12$.

▶ Show solution

A growth curve through $(0, 1)$ and $(1, 4)$, approaching the $x$-axis on the left.

$y$-intercept $(0, 1)$; asymptote $y = 0$.

$4^{-1/2} = \dfrac{1}{\sqrt 4} = \dfrac12$.

Question 2

Write down the exact values of (a) $e^0$ (b) $\ln 1$ (c) $\ln e$ (d) $e^{\ln 5}$ (e) $\ln\left(e^{3}\right)$.

▶ Show solution

(a) $1$ (b) $0$ (c) $1$ (d) $5$ (e) $3$

(d) and (e) use the fact that $e^x$ and $\ln x$ undo each other.

Question 3

Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ for (a) $y = e^{5x}$ (b) $y = 3e^{-2x}$ (c) $y = e^{x/2}$.

▶ Show solution

(a) $5e^{5x}$

(b) $3 \times (-2)e^{-2x} = -6e^{-2x}$

Question 4

Find the gradient of $y = 2e^{3x}$ at the point $(0, 2)$.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6e^{3x}$. At $x = 0$: $6e^0 = 6$.

(Check with $\frac{\mathrm{d}y}{\mathrm{d}x} = ky = 3 \times 2 = 6$ ✓)

Question 5

The curve $y = e^{2x} - 4x$. (a) Find the gradient where $x = 0$. (b) Find the exact $x$-coordinate of the point where the gradient is zero.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2e^{2x} - 4$

(a) At $x = 0$: $2 - 4 = -2$.

(b) $2e^{2x} = 4 \Rightarrow e^{2x} = 2 \Rightarrow 2x = \ln 2 \Rightarrow x = \tfrac12\ln 2$ $(\approx 0.347)$.

Question 6

Sketch $y = 3 - e^{-x}$, giving the equation of the asymptote and the exact coordinates of the points where the curve meets the axes.

▶ Show solution

As $x \to \infty$, $e^{-x} \to 0$, so $y \to 3$: asymptote $y = 3$.

$x = 0$: $y = 3 - 1 = 2$, so $(0, 2)$.

$y = 0$: $e^{-x} = 3 \Rightarrow -x = \ln 3 \Rightarrow x = -\ln 3$, so $(-\ln 3,\, 0)$, about $(-1.10, 0)$.

The curve rises from very negative values on the left, crosses at $(-\ln 3, 0)$ and $(0, 2)$, and levels off below $y = 3$.

Question 7

Sketch $y = \ln(x - 1)$, stating the equation of the asymptote and where it crosses the $x$-axis. Explain why it has no $y$-intercept.

▶ Show solution

$y = \ln x$ translated $1$ unit right: asymptote $x = 1$.

$\ln(x-1) = 0 \Rightarrow x - 1 = 1 \Rightarrow x = 2$: crosses at $(2, 0)$.

At $x = 0$ we would need $\ln(-1)$, which is undefined — the curve only exists for $x > 1$, so it never reaches the $y$-axis.

Question 8

Find the exact coordinates of the point on $y = e^x$ where the gradient is $5$, and the equation of the tangent there.

▶ Show solution

Gradient $= e^x = 5 \Rightarrow x = \ln 5$, and $y = e^{\ln 5} = 5$. Point $(\ln 5,\, 5)$.

Tangent: $y - 5 = 5(x - \ln 5)$

$y = 5x + 5 - 5\ln 5$

Question 9

Show that the curve $y = e^{-x} + 2x$ has exactly one point where the gradient is zero, and find its exact coordinates.

▶ Show solution

$\dfrac{\mathrm{d}y}{\mathrm{d}x} = -e^{-x} + 2$

Zero when $e^{-x} = 2 \Rightarrow -x = \ln 2 \Rightarrow x = -\ln 2$.

Since $\ln$ gives exactly one value, there is exactly one such point.

$y = e^{\ln 2} + 2(-\ln 2) = 2 - 2\ln 2$.

Point: $(-\ln 2,\; 2 - 2\ln 2)$, about $(-0.693,\; 0.614)$.

Question 10

The number of bacteria in a culture after $t$ hours is modelled by $N = 500e^{0.2t}$.

(a) State the initial number of bacteria. (b) Find $N$ when $t = 5$. (c) Find the rate at which the population is increasing when $t = 5$. (d) Show that $\dfrac{\mathrm{d}N}{\mathrm{d}t} = 0.2N$ and explain what this means. (e) Find the time at which there are $2000$ bacteria.

▶ Show solution

(a) $t = 0$: $N = 500e^0 = 500$.

(b) $N = 500e^{1} \approx 1359$ bacteria.

(c) $\dfrac{\mathrm{d}N}{\mathrm{d}t} = 500 \times 0.2e^{0.2t} = 100e^{0.2t}$. At $t = 5$: $100e \approx 272$ bacteria per hour.

(d) $\dfrac{\mathrm{d}N}{\mathrm{d}t} = 100e^{0.2t} = 0.2 \times 500e^{0.2t} = 0.2N$. The rate of growth is proportional to the size of the population: the more bacteria there are, the faster the number grows.

(e) $500e^{0.2t} = 2000 \Rightarrow e^{0.2t} = 4 \Rightarrow 0.2t = \ln 4 \Rightarrow t = 5\ln 4 \approx 6.93$ hours.

Exponential Functions and e · A Level Maths Year 1 · Created with MathJax