The functions aˣ and log_b x and their graphs

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This section builds the two functions the whole unit rests on: the exponential function $a^x$ and the logarithm $\log_b x$ .

It uses the index laws and graph work from International GCSE Mathematics.

Later in this course you meet these functions again inside series (finding the least number of terms), in straight-line graphs, and in calculus, where $e^{ax}$ is differentiated and integrated.

You need to know the functions $a^x$ and $\log_b x$ (where $b$ is a natural number greater than one).

A knowledge of the shape of the graphs of $a^x$ and $\log_b x$ is expected, but not a formal expression for the gradient.

How the marks work

M marks are for a correct method applied to this question, with your own values in. An arithmetic slip keeps the M.

A marks are for a correct answer or line. An A mark needs the M mark before it.

B marks are for a correct fact or value that needs no method, such as an asymptote or $\log_2 8 = 3$ .

dM1 means a method mark that depends on the M mark before it (ddM1: on the two before it). Lose the first and you lose this one too.

cso means "correct solution only". It is the last mark of a printed result, which is starred (A1* cso or A1cso): every line must be right.

ft (follow through) lets you score with your own earlier wrong value. oe means "or equivalent". awrt means "answer which rounds to", so $2.53$ earns "awrt $2.5$ ".

Four rules for this course. A quadratic's M needs visible factorising, the formula with values in, or a completed square. A calculator root earns the M only after a correct three-term quadratic is shown. "Exact" means exact on every line: one decimal and exactness is lost for good. A misread number costs two A marks, but the M marks are kept.

The idea: repeated multiplying

Start with a doubling pattern.

A sheet of paper folded in half has 2 layers; fold again and there are 4, then 8, then 16.

After $x$ folds there are $2^x$ layers.

$x$

$-2$

$-1$

$0$

$1$

$2$

$3$

$4$

$2^x$

$\frac{1}{4}$

$\frac{1}{2}$

$1$

$2$

$4$

$8$

$16$

Read the table from right to left: every step left halves the value.

That is why $2^0 = 1$ and $2^{-1} = \frac{1}{2}$ .

The index laws you know from International GCSE Mathematics say the same thing: $a^0 = 1$ , $a^{-n} = \frac{1}{a^n}$ and $a^{\frac{1}{n}} = \sqrt[n]{a}$ .

So $a^x$ has a value for every real $x$ , not only whole numbers.

For example $2^{0.5} = \sqrt{2} \approx 1.41$ .

A function of the form $y = a^x$ , with the variable in the index, is called an exponential function.

The number $a$ must be positive.

A negative base fails: $(-4)^{\frac{1}{2}}$ would be $\sqrt{-4}$ , which has no real value.

The shape of $y = a^x$

The graph below shows $y = 2^x$ , $y = 3^x$ and $y = \left(\frac{1}{2}\right)^x$ , which is labelled $y = 2^{-x}$ because $\left(\frac{1}{2}\right)^x = 2^{-x}$ .

Three exponential curves on one set of axes. The curves y = 2^x and y = 3^x rise from left to right, staying just above the x-axis on the left, and y = 3^x is the steeper of the two for positive x. The curve y = (1/2)^x, labelled y = 2^(−x), falls from left to right and is the mirror image of y = 2^x in the y-axis. All three pass through the point (0, 1), and none of them touches or crosses the x-axis, which is an asymptote to each.

Every curve $y = a^x$ (with $a > 0$ ) has these features.

Learn them as a list, because a sketch is marked on them.

It passes through $(0, 1)$ , since $a^0 = 1$ for every base.

It is always positive. However far left you go, $2^x$ only gets closer to $0$ ; it never reaches it. So the $x$ -axis, $y = 0$ , is an asymptote.

If $a > 1$ it rises from left to right, and the larger $a$ is, the steeper it is for $x > 0$ .

If $0 < a < 1$ it falls. In fact $\left(\frac{1}{2}\right)^x = 2^{-x}$ , so $y = \left(\frac{1}{2}\right)^x$ is the reflection of $y = 2^x$ in the $y$ -axis.

A curve that rises like $2^x$ models growth where a quantity is multiplied by the same factor in each equal time step: money with compound interest, a population of bacteria.

A falling curve like $\left(\frac{1}{2}\right)^x$ models decay: a cooling drink, a medicine leaving the blood.

You solve problems on these in the section on change of base.

The idea of a logarithm: "the power you need"

The table above answers "what is $2^3$ ?".

A logarithm answers the reverse question: "to what power must I raise 2 to get 8?".

The answer is 3, and we write

\log_2 8 = 3.

In general, for a base $b$ ,

\log_b x = n \quad\text{means exactly the same as}\quad b^n = x.

Say it aloud as "the log of $x$ to base $b$ is the power you need on $b$ to make $x$ ".

This one sentence is the definition, and almost every log question in this unit is solved by going back to it.

Index form

Log form

$3^4 = 81$

$\log_3 81 = 4$

$10^{-2} = 0.01$

$\log_{10} 0.01 = -2$

$5^0 = 1$

$\log_5 1 = 0$

$16^{\frac{1}{2}} = 4$

$\log_{16} 4 = \frac{1}{2}$

Two things follow at once.

The log of a negative number or of zero does not exist. No power of a positive base is negative or zero, because $b^n > 0$ for every $n$ . So $\log_2(-8)$ and $\log_2 0$ have no value. The number inside the log, called the argument, must be positive.

A log can be negative or a fraction. $\log_2 \frac{1}{8} = -3$ because $2^{-3} = \frac{1}{8}$ . A negative log means the argument is between 0 and 1, not that the argument is negative.

Calculator keys. $\log_{10} x$ is often written $\lg x$ , and the calculator's log key gives it.

The ln key gives $\ln x$ , the logarithm to base $e$ , where $e \approx 2.718$ is a special number you use in calculus.

Any other base is found with change of base, later in this unit.

The logarithm was first published by John Napier in 1614, as a way to turn long multiplications into additions.

The shape of $y = \log_b x$

Because $\log_2 x$ undoes $2^x$ , its graph is the graph of $2^x$ with $x$ and $y$ swapped.

Swapping the coordinates reflects a graph in the line $y = x$ , as the graph below shows (both axes use the same scale).

The curve y = 2^x and the curve y = log_2 x drawn on equal scales with the dashed line y = x. Each curve is the reflection of the other in y = x. The point (0, 1) on y = 2^x reflects to (1, 0) on y = log_2 x, and (2, 4) reflects to (4, 2). The exponential curve has the x-axis as an asymptote; the log curve has the y-axis as an asymptote and exists only for x greater than 0.

So every curve $y = \log_b x$ (with $b$ a natural number greater than one) has the reflected features:

It passes through $(1, 0)$ , since $b^0 = 1$ , and through $(b, 1)$ , since $b^1 = b$ .

It exists only for $x > 0$ , and the $y$ -axis, $x = 0$ , is an asymptote.

It is negative for $0 < x < 1$ and positive for $x > 1$ .

It always rises, but more and more slowly. $\log_2 x$ only reaches 10 at $x = 1024$ .

The next graph shows three bases together.

The larger the base, the flatter the curve.

The curves y = log_2 x, y = log_3 x and y = lg x for x from 0.1 to 10. All three pass through (1, 0), are negative for x between 0 and 1, and fall steeply towards the y-axis, which none of them touches. Each meets the dashed line y = 1 at x equal to its base: (2, 1), (3, 1) and (10, 1). The larger the base, the more slowly the curve rises.

Moving the curves. Graph transformations from International GCSE Mathematics work as usual.

$y = 2^x + 3$ is $y = 2^x$ moved up 3, so its asymptote is $y = 3$ .

$y = \log_3(x + 3)$ is $y = \log_3 x$ moved 3 to the left, so its asymptote is $x = -3$ .

When you sketch, move the asymptote first; then find where the new curve crosses the axes.

Why there is no gradient formula here. You are not asked for a formula for the gradient of these curves in this unit.

In calculus you learn the derivative of $e^{ax}$ ; the gradient of $\log_b x$ is not on this course.

Key terms

Exponential function: a function of the form $a^x$ , with $a > 0$ , where the variable is the index.

Logarithm: $\log_b x$ is the power to which the base $b$ must be raised to give $x$ ; $\log_b x = n$ means $b^n = x$ .

Argument of a logarithm: the number or expression the log is taken of; it must be positive.

Asymptote: a line that a curve gets closer and closer to but never meets; $y = 0$ for $y = a^x$ and $x = 0$ for $y = \log_b x$ .

lg and ln: $\lg x$ is $\log_{10} x$ ; $\ln x$ is the natural logarithm, $\log_e x$ .

Worked example 1: values from the definition

Without using a calculator, find the value of (a) $\log_4 8$ (b) $\lg 0.001$ (c) $\log_{27} \frac{1}{3}$ (3)

1. (a) $4 = 2^2$ and $8 = 2^3$ , so $4^{\frac{3}{2}} = (2^2)^{\frac{3}{2}} = 2^3 = 8$ , giving $\log_4 8 = \frac{3}{2}$ B1

2. (b) $0.001 = 10^{-3}$ , so $\lg 0.001 = -3$ B1

3. (c) $27^{\frac{1}{3}} = 3$ , so $27^{-\frac{1}{3}} = \frac{1}{3}$ , giving $\log_{27} \frac{1}{3} = -\frac{1}{3}$ B1

Answer: $\frac{3}{2}$ , $-3$ , $-\frac{1}{3}$ .

Check each by turning it back into index form: $4^{1.5} = 8$ , $10^{-3} = 0.001$ , $27^{-1/3} = \frac{1}{3}$ .

When base and argument are both powers of one number, write both as powers of it first.

Figure 1

Diagram NOT accurately drawn

A sketch of y = log_3 (x + 3). The curve rises from the bottom of the frame close to the dashed vertical asymptote x = −3, crosses the x-axis at (−2, 0), crosses the y-axis at (0, 1) and keeps rising ever more slowly to the right. It never reaches the asymptote.

Worked example 2: a "show that", then a sketch

The curve $C$ has equation $y = \log_3 (x + k)$ , where $k$ is a constant.

The point $(6, 2)$ lies on $C$ .

(a) Show that $k = 3$ (2)

(b) Sketch $C$ , showing the coordinates of any points where $C$ meets the coordinate axes and the equation of the asymptote. (4)

1. (a) $2 = \log_3 (6 + k)$ , so $6 + k = 3^2$ M1 (definition: log form to index form)

2. $6 + k = 9$ , so $k = 3$ * A1cso

3. (b) The asymptote is where the argument is zero: $x + 3 = 0$ , so $x = -3$ B1

4. On the $x$ -axis, $y = 0$ : $\log_3 (x + 3) = 0 \Rightarrow x + 3 = 3^0 = 1 \Rightarrow (-2, 0)$ B1

5. On the $y$ -axis, $x = 0$ : $y = \log_3 3 = 1 \Rightarrow (0, 1)$ B1

6. A rising curve, to the right of $x = -3$ only, getting ever closer to the asymptote and never crossing it, as in Figure 1 B1

Answer: $k = 3$ ; the curve in Figure 1.

In part (a) the line $6 + k = 9$ must be seen: writing " $k = 3$ " alone gives no method, and the A1 is lost.

Worked example 3: drawing a line to solve an equation

(a) Complete the table of values for $y = 2^{x-1}$ , giving your values to 2 decimal places. (2)

$x$

$-1$

$-0.5$

$0$

$0.5$

$1$

$1.5$

$2$

$2.5$

$3$

$y$

$0.25$

$0.5$

$1$

$2$

$4$

(b) Draw the graph of $y = 2^{x-1}$ for $-1 \leqslant x \leqslant 3$ (2)

(c) By drawing a suitable straight line on your graph, obtain an estimate of the root of $2^x + 3x - 9 = 0$ , giving your answer to one decimal place. (4)

1. (a) $0.35$ , $0.71$ , $1.41$ , $2.83$ (all four correct) B2 (B1 for two or three correct)

2. (b) All nine points plotted to within half a small square B1

3. A smooth curve through all the points, with no straight sections B1

4. (c) Make one side the printed curve. $2^x = 2 \times 2^{x - 1}$ , so $2 \times 2^{x-1} + 3x - 9 = 0$ M1

5. $2^{x-1} = \frac{9 - 3x}{2}$ , so the line is $y = 4.5 - 1.5x$ A1

6. The line $y = 4.5 - 1.5x$ drawn, through $(-1, 6)$ and $(3, 0)$ , with its equation written beside it M1

7. Where the line meets the curve, $x = 1.8$ A1 (awrt 1.8)

Answer: $x \approx 1.8$ .

Check: $2^{1.8} + 3(1.8) - 9 = 3.48 + 5.4 - 9 = -0.12$ , close to zero, and at $x = 1.85$ the left side is already positive, so the root lies just above 1.8.

The nine plotted points of y = 2^(x − 1) from x = −1 to x = 3 joined by a smooth rising curve, and the straight line y = 4.5 − 1.5x falling from (−1, 6) to (3, 0). They cross once, and a dashed guide drops from the crossing to the x-axis at about x = 1.8.

The graph above is the finished answer to Worked example 3.

The method is the same for every graphical item: rearrange the equation you are given until one side is exactly the curve already drawn, and whatever is left is the line.

The algebra that produces the line earns the first M1 A1; without it, roots read from a calculator earn nothing.

Worked example 4: finding the constants of an exponential curve

The curve with equation $y = ab^x$ , where $a$ and $b$ are positive constants, passes through the points $(1, 6)$ and $(3, 54)$ .

Find the value of $a$ and the value of $b$ .

(4)

1. $ab = 6$ and $ab^3 = 54$ B1 (both points substituted)

2. Divide: $\frac{ab^3}{ab} = \frac{54}{6}$ , so $b^2 = 9$ M1

3. $b = 3$ (as $b > 0$ ) A1

4. $3a = 6$ , so $a = 2$ A1

Answer: $a = 2$ , $b = 3$ , so $y = 2 \times 3^x$ .

Check: $2 \times 3^3 = 54$ .

Dividing the two equations removes $a$ in one step, the same trick you will use for geometric series in a later unit.

Exam tip

Give exactly the accuracy the stem asks for, no more and no fewer figures.

A table "to 2 decimal places" needs $0.71$ , not $0.7$ or $0.707$ ; a root read from a hand-drawn graph "to one decimal place" needs $1.8$ .

A graph reading given to more decimal places than asked loses the A mark.

Common mistakes

Taking the log of a negative number or zero. No power of a positive base is negative or zero, so $\log_2(-8)$ does not exist. A negative log, such as $\log_2 \frac{1}{8} = -3$ , is fine: it means the argument is between 0 and 1.

Drawing $y = a^x$ crossing the $x$ -axis. The curve is always above it. Draw it getting close on one side, never touching.

Too few steps in a "show that". In Worked example 2, jumping from the point to " $k = 3$ " loses the method mark. Write the index-form line, then the value, and finish with the printed result.

A straight line read without its equation. In a graphical item, write the rearrangement and the line's equation before you draw it. A root read with no line equation scores nothing for method.

Mixing up the asymptotes. $y = a^x$ has the horizontal asymptote $y = 0$ ; $y = \log_b x$ has the vertical asymptote $x = 0$ . Move them with the curve when it is shifted.

Check your understanding

1. Without using a calculator, find the value of
(a) (i) $\log_2 \frac{1}{16}$
(ii) $\log_8 16$ (2)
(b) Find the value of $p$ for which $\log_p 125 = \frac{3}{2}$ (2) Total [4]

2. The curve $C$ has equation $y = 5 - 2^x$ .
(a) Write down the equation of the asymptote to $C$ . (1)
(b) Find the coordinates of the points where $C$ meets the coordinate axes, giving exact values. (3)
(c) Sketch $C$ , showing the asymptote with its equation and each axis intercept with its coordinates. (3) Total [7]

3. (a) A student writes: " $\log_2 x = -3$ , so $x = -8$ ." Find the correct value of $x$ . (1) The curve $C$ has equation $y = \log_7 (x - 4)$ .
(b) Find the set of values of $x$ for which $y$ has a value. (1)
(c) Find the coordinates of the point where $C$ crosses the $x$ -axis, and the equation of the asymptote to $C$ . (2) Total [4]

4. The curve with equation $y = ab^x$ , where $a$ and $b$ are positive constants, passes through the points $(2, 12)$ and $(5, 96)$ .
(a) Find the value of $a$ and the value of $b$ . (4)
(b) Find the value of $x$ for which $y = 384$ (2) Total [6]

5. (a) For $y = \log_2 (x + 1)$ , complete the table, giving each missing value to 2 decimal places.

$x$

$0$

$0.5$

$1$

$2$

$3$

$4$

$5$

$6$

$7$

$y$

$0$

$1$

$2$

$3$

(2)
(b) Draw the graph of $y = \log_2 (x + 1)$ for $0 \leqslant x \leqslant 7$ (2)
(c) By drawing a suitable straight line on your graph, obtain an estimate of the solution of $x + 1 = 2^{\frac{8 - x}{3}}$ for $0 \leqslant x \leqslant 7$ , giving your answer to one decimal place. (4) Total [8]

6. The curves $y = 2^x$ and $y = 8^{x - 1}$ meet at the point $P$ .
(a) Show that the $x$ -coordinate of $P$ is $\frac{3}{2}$ (3)
(b) Find the $y$ -coordinate of $P$ , giving your answer in the form $a\sqrt{b}$ , where $a$ and $b$ are integers. (2) Total [5]

7. (a) Sketch, on the same axes, the curves $y = 3^x$ and $y = \log_3 x$ , showing the coordinates of any point where each curve meets a coordinate axis. (3)
(b) Hence find the number of real solutions of the equation $3^x = \log_3 x$ (1) Total [4]

Answers

1. (a)(i) $2^{-4} = \frac{1}{16}$ , so $-4$ B1
(ii) $8 = 2^3$ , $16 = 2^4$ , so $8^{\frac{4}{3}} = 16$ , giving $\frac{4}{3}$ B1
(b) $p^{\frac{3}{2}} = 125$ M1; $p = 125^{\frac{2}{3}} = 25$ A1

2. (a) $y = 5$ B1
(b) $x = 0$ : $y = 5 - 1 = 4$ , so $(0, 4)$ B1; $y = 0$ : $2^x = 5$ M1; $x = \log_2 5$ , so $(\log_2 5, 0)$ A1
(c) A curve falling from left to right, approaching $y = 5$ from below on the left B1; the asymptote drawn and labelled $y = 5$ B1; $(0, 4)$ and $(\log_2 5, 0)$ labelled (about $(2.32, 0)$ ) B1

3. (a) $x = 2^{-3} = \frac{1}{8}$ B1 (a log can be negative; the argument cannot)
(b) The argument must be positive: $x - 4 > 0$ , so $x > 4$ B1
(c) $y = 0$ : $x - 4 = 7^0 = 1$ , so $(5, 0)$ B1; asymptote $x = 4$ B1

4. (a) $ab^2 = 12$ and $ab^5 = 96$ B1; dividing, $b^3 = 8$ M1; $b = 2$ A1; $4a = 12$ , $a = 3$ A1
(b) $3 \times 2^x = 384 \Rightarrow 2^x = 128$ M1; $x = 7$ A1

5. (a) $0.58$ , $1.58$ , $2.32$ , $2.58$ , $2.81$ (all correct) B2 (B1 for three or four correct)
(b) Points plotted to within half a small square B1; a smooth curve through them B1
(c) $x + 1 = 2^{\frac{8-x}{3}}$ means $\log_2 (x + 1) = \frac{8 - x}{3}$ (definition of a log) M1; the line $y = \frac{8 - x}{3}$ A1; the line drawn, through $(0, \frac{8}{3})$ and $(5, 1)$ M1; $x = 2.5$ A1 (awrt 2.5)

6. (a) $8^{x-1} = (2^3)^{x - 1} = 2^{3x - 3}$ M1; $2^x = 2^{3x - 3} \Rightarrow x = 3x - 3$ M1; $x = \frac{3}{2}$ * A1cso
(b) $y = 2^{\frac{3}{2}} = 2 \times 2^{\frac{1}{2}}$ M1; $y = 2\sqrt{2}$ A1

7. (a) $y = 3^x$ rising through $(0, 1)$ , above the $x$ -axis B1; $y = \log_3 x$ rising through $(1, 0)$ , to the right of the $y$ -axis B1; the two curves are reflections in $y = x$ and do not meet B1
(b) None: $y = 3^x$ lies above the line $y = x$ and $y = \log_3 x$ lies below it, so the curves never meet B1

The functions aˣ and log_b x and their graphs

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