Sec 3 · Additional Mathematics

Exponential and logarithmic functions

Practise exponential and logarithmic functions with step-by-step help. Free account.

1. Exponential equations

Same base: equate the powers

\displaystyle 2^{x + 1} = 32

e.g.

\displaystyle 2^{x + 1} = 2^{5}

\displaystyle x + 1 = 5

\displaystyle x = 4

Make the bases the same

\displaystyle 4^{x} = 8^{x - 1}

e.g.

\displaystyle 2^{2x} = 2^{3(x - 1)}

\displaystyle 2x = 3x - 3

\displaystyle x = 3

Hidden quadratic: let y = 3ˣ

\displaystyle 3^{2x} - 4(3^{x}) + 3 = 0

e.g.

\displaystyle y^{2} - 4y + 3 = 0

\displaystyle (y - 1)(y - 3) = 0

\displaystyle y = 1 \text{ or } y = 3

$3^{x} = 1$ , so $x = 0$
$3^{x} = 3$ , so $x = 1$

📖 Worked example

Solve $3^{2x} - 4(3^x) + 3 = 0$ .

Step 1 of 4

Write 3²ˣ as (3ˣ)²: then it is a quadratic in 3ˣ

3^{2x} = (3^{x})^{2}

(3^{x})^2 - 4(3^{x}) + 3 = 0

Practise this →

2. Logarithms and their laws

Name	Rule	Example
definition	$y = a^{x}$ means $x = \log_{a} y$ ( $a > 0$ , $a \neq 1$ , $y > 0$ )	$8 = 2^{3}$ means $3 = \log_{2} 8$
	$\log_{a} 1 = 0$	$\log_{5} 1 = 0$
	$\log_{a} a = 1$	$\log_{5} 5 = 1$
product law	$\log_{a} mn = \log_{a} m + \log_{a} n$	$\log_{2} 12 = \log_{2} 4 + \log_{2} 3$
quotient law	$\log_{a} \dfrac{m}{n} = \log_{a} m - \log_{a} n$	$\log_{2} \dfrac{24}{3} = \log_{2} 24 - \log_{2} 3$
power law	$\log_{a} m^{k} = k\log_{a} m$	$\log_{2} 9 = 2\log_{2} 3$
change of base	$\log_{b} N = \dfrac{\log_{a} N}{\log_{a} b}$	$\log_{4} 8 = \dfrac{\log_{2} 8}{\log_{2} 4}$
common log	$\lg x = \log_{10} x$	$\lg 1000 = 3$
natural log	$\ln x = \log_{e} x$	$\ln e^{2} = 2$

Combine into one log, then evaluate

\displaystyle \log_{2} 24 - \log_{2} 3

e.g.

\displaystyle \log_{2} 24 - \log_{2} 3

\displaystyle = \log_{2} 8

\displaystyle = 3

Change of base

\displaystyle \log_{4} 8

e.g.

\displaystyle \log_{4} 8

\displaystyle = \dfrac{\log_{2} 8}{\log_{2} 4}

\displaystyle = \dfrac{3}{2}

📖 Worked example

Without using a calculator, find the value of $\log_{4} 8$ .

Step 1 of 3

Change of base to 2: the log of the number ÷ the log of the base

\log_{4} 8 = \dfrac{\log_{2} 8}{\log_{2} 4}

Practise this →

3. Graphs of exponential and logarithmic functions

y = aˣ and y = logₐ x

\displaystyle y = 2^{x}

y = 2 to the x through (0, 1) and (1, 2); y = log base 2 of x through (1, 0) and (2, 1); they are reflections in the dashed line y = x.

passes through $(0, 1)$ ; $y > 0$ for every $x$
$y = \log_{2} x$ is its reflection in $y = x$ : through $(1, 0)$ , only for $x > 0$

y = eˣ and y = ln x

\displaystyle e = 2.718\ldots

y = e to the x through (0, 1) and y = ln x through (1, 0), reflections of each other in y = x.

$y = e^{x}$ through $(0, 1)$ ; $y = \ln x$ through $(1, 0)$

Curve	Asymptote	Shape
$y = c + ae^{x}$	$y = c$	$a > 0$ : rises from $y = c$ ; $a < 0$ : falls from it
$y = c + ae^{-x}$	$y = c$	$a > 0$ : falls to $y = c$ ; $a < 0$ : rises to it
$y = c + a\ln(x + h)$	$x = -h$	only for $x > -h$ ; $a > 0$ : rises, $a < 0$ : falls

Where it meets an axis

\displaystyle y = e^{x} - 2

e.g.

\displaystyle e^{x} - 2 = 0

\displaystyle e^{x} = 2

\displaystyle x = \ln 2

Number of solutions: one side is the curve, the other a line to draw

\displaystyle e^{x} + 2x = 0

$e^{x} - 2 = -2x - 2$ : draw $y = -2x - 2$ on the graph of $y = e^{x} - 2$
they meet once: $1$ solution

📖 Worked example

State the equation of the asymptote of the curve $y = \ln(x - 2) - 1$ .

Step 1 of 1

ln is defined only for positive numbers: the asymptote is where the number inside reaches 0

x - 2 = 0

x = 2

📖 Worked example

Find the exact $x$ -coordinate of the point where the curve $y = e^{x} - 2$ meets the $x$ -axis.

Step 1 of 2

On the x-axis, y = 0: make e to the power the subject

e^{x} - 2 = 0

e^{x} = 2

📖 Worked example

The diagram shows the curve $y = e^{x} - 2$ and its asymptote. By drawing a suitable straight line on the diagram, state the number of solutions of the equation $e^{x} + 2x = 0$ .

The curve y = e^x - 2 with its asymptote dashed.

Step 1 of 2

Rearrange so that the left side is the curve's own equation: the right side is then the line to draw

e^{x} - 2 = -2x - 2

Draw the line

y = -2x - 2

Practise this →

4. Logarithmic equations

One log: write in index form

\displaystyle \log_{2}(x + 1) = 3

e.g.

\displaystyle \log_{2}(x + 1) = 3

\displaystyle x + 1 = 2^{3}

\displaystyle x = 7

Natural log

\displaystyle \ln(2x) = 3

e.g.

\displaystyle \ln(2x) = 3

\displaystyle 2x = e^{3}

\displaystyle x = 10.0\text{ (3 s.f.)}

📖 Worked example

Solve $\log_{2} x + \log_{2} (x - 2) = 3$ .

Step 1 of 5

Which law of logarithms combines the two logs into one?

Product law:

\log_{a} m + \log_{a} n = \log_{a} mn

Different bases: change of base first

\displaystyle \log_{4} x = \dfrac{\log_{2} x}{2}

$\log_{2} x + \log_{4} x = -3$
$\dfrac{3}{2}\log_{2} x = -3$ , so $\log_{2} x = -2$
$x = 2^{-2}$ , which is $\dfrac{1}{4}$

logₓ a = 1 ÷ logₐ x: a quadratic in u = logₐ x

\displaystyle 2\log_{x} 3 + \log_{3} x = 3

$\dfrac{2}{u} + u = 3$ , so $u^{2} - 3u + 2 = 0$
$u = 1$ or $u = 2$
$x = 3$ or $x = 9$

📖 Worked example

Solve the equation $\log_{2} x + \log_{4} x = -3$ .

Step 1 of 3

Change of base: write log₄ x with base 2

\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4}

= \dfrac{\log_{2} x}{2}

📖 Worked example

Solve the equation $2\log_{x} 3 + \log_{3} x = 3$ .

Step 1 of 4

Change of base: logₓ 3 = 1 ÷ log₃ x. Let u = log₃ x

\log_{x} 3 = \dfrac{1}{\log_{3} x}

u + \dfrac{2}{u} = 3

Practise this →

5. Solving equations with logarithms

Take lg of both sides

\displaystyle 3^{x} = 20

e.g.

\displaystyle 3^{x} = 20

\displaystyle x = \dfrac{\lg 20}{\lg 3}

\displaystyle x = 2.73\text{ (3 s.f.)}

Base e: take ln of both sides

\displaystyle e^{x} = 5

e.g.

\displaystyle e^{x} = 5

\displaystyle x = \ln 5

\displaystyle x = 1.61\text{ (3 s.f.)}

📖 Worked example

Solve $3^{2x - 1} = 20$ , giving your answer correct to 3 significant figures.

Step 1 of 3

Take lg of both sides; the power law brings the power down

\lg(3^{2x - 1}) = \lg 20

(2x - 1) \lg 3 = \lg 20

Practise this →

6. Exponential and logarithmic models

Decay: N = 500e^(−0.2t). Find t when N = 100

\displaystyle N = 500e^{-0.2t}

e.g.

\displaystyle 500e^{-0.2t} = 100

\displaystyle e^{-0.2t} = \dfrac{1}{5}

\displaystyle -0.2t = \ln \dfrac{1}{5}

\displaystyle t = 8.05\text{ (3 s.f.)}

the start value: $t = 0$ gives $N = 500$
when $t = 3$ : $N = 500e^{-0.6}$ , which is $274$ (3 s.f.)

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Sec 3 · Additional Mathematics

Exponential and logarithmic functions

Practise exponential and logarithmic functions with step-by-step help. Free account.

Start free Worksheet (PDF)

1. Exponential equations

Same base: equate the powers

\displaystyle 2^{x + 1} = 32

e.g.

\displaystyle 2^{x + 1} = 2^{5}

\displaystyle x + 1 = 5

\displaystyle x = 4

Make the bases the same

\displaystyle 4^{x} = 8^{x - 1}

e.g.

\displaystyle 2^{2x} = 2^{3(x - 1)}

\displaystyle 2x = 3x - 3

\displaystyle x = 3

Hidden quadratic: let y = 3ˣ

\displaystyle 3^{2x} - 4(3^{x}) + 3 = 0

e.g.

\displaystyle y^{2} - 4y + 3 = 0

\displaystyle (y - 1)(y - 3) = 0

\displaystyle y = 1 \text{ or } y = 3

$3^{x} = 1$ , so $x = 0$
$3^{x} = 3$ , so $x = 1$

📖 Worked example

Solve $3^{2x} - 4(3^x) + 3 = 0$ .

Step 1 of 4

Write 3²ˣ as (3ˣ)²: then it is a quadratic in 3ˣ

3^{2x} = (3^{x})^{2}

(3^{x})^2 - 4(3^{x}) + 3 = 0

Practise this →

2. Logarithms and their laws

Name	Rule	Example
definition	$y = a^{x}$ means $x = \log_{a} y$ ( $a > 0$ , $a \neq 1$ , $y > 0$ )	$8 = 2^{3}$ means $3 = \log_{2} 8$
	$\log_{a} 1 = 0$	$\log_{5} 1 = 0$
	$\log_{a} a = 1$	$\log_{5} 5 = 1$
product law	$\log_{a} mn = \log_{a} m + \log_{a} n$	$\log_{2} 12 = \log_{2} 4 + \log_{2} 3$
quotient law	$\log_{a} \dfrac{m}{n} = \log_{a} m - \log_{a} n$	$\log_{2} \dfrac{24}{3} = \log_{2} 24 - \log_{2} 3$
power law	$\log_{a} m^{k} = k\log_{a} m$	$\log_{2} 9 = 2\log_{2} 3$
change of base	$\log_{b} N = \dfrac{\log_{a} N}{\log_{a} b}$	$\log_{4} 8 = \dfrac{\log_{2} 8}{\log_{2} 4}$
common log	$\lg x = \log_{10} x$	$\lg 1000 = 3$
natural log	$\ln x = \log_{e} x$	$\ln e^{2} = 2$

Combine into one log, then evaluate

\displaystyle \log_{2} 24 - \log_{2} 3

e.g.

\displaystyle \log_{2} 24 - \log_{2} 3

\displaystyle = \log_{2} 8

\displaystyle = 3

Change of base

\displaystyle \log_{4} 8

e.g.

\displaystyle \log_{4} 8

\displaystyle = \dfrac{\log_{2} 8}{\log_{2} 4}

\displaystyle = \dfrac{3}{2}

📖 Worked example

Without using a calculator, find the value of $\log_{4} 8$ .

Step 1 of 3

Change of base to 2: the log of the number ÷ the log of the base

\log_{4} 8 = \dfrac{\log_{2} 8}{\log_{2} 4}

Practise this →

3. Graphs of exponential and logarithmic functions

y = aˣ and y = logₐ x

\displaystyle y = 2^{x}

passes through $(0, 1)$ ; $y > 0$ for every $x$
$y = \log_{2} x$ is its reflection in $y = x$ : through $(1, 0)$ , only for $x > 0$

y = eˣ and y = ln x

\displaystyle e = 2.718\ldots

$y = e^{x}$ through $(0, 1)$ ; $y = \ln x$ through $(1, 0)$

Curve	Asymptote	Shape
$y = c + ae^{x}$	$y = c$	$a > 0$ : rises from $y = c$ ; $a < 0$ : falls from it
$y = c + ae^{-x}$	$y = c$	$a > 0$ : falls to $y = c$ ; $a < 0$ : rises to it
$y = c + a\ln(x + h)$	$x = -h$	only for $x > -h$ ; $a > 0$ : rises, $a < 0$ : falls

Where it meets an axis

\displaystyle y = e^{x} - 2

e.g.

\displaystyle e^{x} - 2 = 0

\displaystyle e^{x} = 2

\displaystyle x = \ln 2

Number of solutions: one side is the curve, the other a line to draw

\displaystyle e^{x} + 2x = 0

$e^{x} - 2 = -2x - 2$ : draw $y = -2x - 2$ on the graph of $y = e^{x} - 2$
they meet once: $1$ solution

📖 Worked example

State the equation of the asymptote of the curve $y = \ln(x - 2) - 1$ .

Step 1 of 1

ln is defined only for positive numbers: the asymptote is where the number inside reaches 0

x - 2 = 0

x = 2

📖 Worked example

Find the exact $x$ -coordinate of the point where the curve $y = e^{x} - 2$ meets the $x$ -axis.

Step 1 of 2

On the x-axis, y = 0: make e to the power the subject

e^{x} - 2 = 0

e^{x} = 2

📖 Worked example

The diagram shows the curve $y = e^{x} - 2$ and its asymptote. By drawing a suitable straight line on the diagram, state the number of solutions of the equation $e^{x} + 2x = 0$ .

Step 1 of 2

Rearrange so that the left side is the curve's own equation: the right side is then the line to draw

e^{x} - 2 = -2x - 2

Draw the line

y = -2x - 2

Practise this →

4. Logarithmic equations

One log: write in index form

\displaystyle \log_{2}(x + 1) = 3

e.g.

\displaystyle \log_{2}(x + 1) = 3

\displaystyle x + 1 = 2^{3}

\displaystyle x = 7

Natural log

\displaystyle \ln(2x) = 3

e.g.

\displaystyle \ln(2x) = 3

\displaystyle 2x = e^{3}

\displaystyle x = 10.0\text{ (3 s.f.)}

📖 Worked example

Solve $\log_{2} x + \log_{2} (x - 2) = 3$ .

Step 1 of 5

Which law of logarithms combines the two logs into one?

Product law:

\log_{a} m + \log_{a} n = \log_{a} mn

Different bases: change of base first

\displaystyle \log_{4} x = \dfrac{\log_{2} x}{2}

$\log_{2} x + \log_{4} x = -3$
$\dfrac{3}{2}\log_{2} x = -3$ , so $\log_{2} x = -2$
$x = 2^{-2}$ , which is $\dfrac{1}{4}$

logₓ a = 1 ÷ logₐ x: a quadratic in u = logₐ x

\displaystyle 2\log_{x} 3 + \log_{3} x = 3

$\dfrac{2}{u} + u = 3$ , so $u^{2} - 3u + 2 = 0$
$u = 1$ or $u = 2$
$x = 3$ or $x = 9$

📖 Worked example

Solve the equation $\log_{2} x + \log_{4} x = -3$ .

Step 1 of 3

Change of base: write log₄ x with base 2

\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4}

= \dfrac{\log_{2} x}{2}

📖 Worked example

Solve the equation $2\log_{x} 3 + \log_{3} x = 3$ .

Step 1 of 4

Change of base: logₓ 3 = 1 ÷ log₃ x. Let u = log₃ x

\log_{x} 3 = \dfrac{1}{\log_{3} x}

u + \dfrac{2}{u} = 3

Practise this →

5. Solving equations with logarithms

Take lg of both sides

\displaystyle 3^{x} = 20

e.g.

\displaystyle 3^{x} = 20

\displaystyle x = \dfrac{\lg 20}{\lg 3}

\displaystyle x = 2.73\text{ (3 s.f.)}

Base e: take ln of both sides

\displaystyle e^{x} = 5

e.g.

\displaystyle e^{x} = 5

\displaystyle x = \ln 5

\displaystyle x = 1.61\text{ (3 s.f.)}

📖 Worked example

Solve $3^{2x - 1} = 20$ , giving your answer correct to 3 significant figures.

Step 1 of 3

Take lg of both sides; the power law brings the power down

\lg(3^{2x - 1}) = \lg 20

(2x - 1) \lg 3 = \lg 20

Practise this →

6. Exponential and logarithmic models

Decay: N = 500e^(−0.2t). Find t when N = 100

\displaystyle N = 500e^{-0.2t}

e.g.

\displaystyle 500e^{-0.2t} = 100

\displaystyle e^{-0.2t} = \dfrac{1}{5}

\displaystyle -0.2t = \ln \dfrac{1}{5}

\displaystyle t = 8.05\text{ (3 s.f.)}

the start value: $t = 0$ gives $N = 500$
when $t = 3$ : $N = 500e^{-0.6}$ , which is $274$ (3 s.f.)

Try a question: Exponential and logarithmic functions

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

Practise exponential and logarithmic functions with step-by-step help

Endless questions from basic to exam level. Get one wrong and see where it went wrong. Free account.

Start free