Mathematics guide, Foundations, step 5 of 33Previous step Next step

Step 5: Exponentials and Logarithms

By the EconTutorials team. 12-minute read with 6 practice questions.

Syllabus9709 P2/P3 2.2, 3.2 · IB SL 1.5, 1.7, 2.9

In 30 seconds

In an exponential function $ y = a^x $ the base is fixed and the index varies. Every such curve passes through $ (0, 1) $ and stays above the asymptote $ y = 0 $: growth if $ a > 1 $, decay if $ 0 < a < 1 $.
A logarithm is an index: $ \log_a x = y $ means $ a^y = x $. The laws $ \log(xy) = \log x + \log y $, $ \log(x/y) = \log x-\log y $ and $ \log x^n = n\log x $ come straight from the index laws.
To solve $ a^x = b $, take logs of both sides: $ x = \dfrac{\ln b}{\ln a} $. The natural logarithm $ \ln x $ is the inverse of $ e^x $, and $ y = Ae^{kt} $ has doubling time $ \dfrac{\ln 2}{k} $.
Exam point: $ \ln 0.9 $ and every log of a number between 0 and 1 is negative, so dividing by it reverses an inequality. Also reject any answer that would need the log of zero or a negative number.

Video lesson coming soonA short animated lesson for this step will appear here.

Exponential functions and their graphs

Many quantities grow by multiplying by the same factor each period: savings earning interest, prices under steady inflation, a country’s output. Others shrink the same way, like a car that loses a fixed percentage of its value each year. These are exponential functions. The question you usually want answered, “how long?”, sits up in the index, and logarithms are the tool that brings it down.

In Step 1 the index was fixed and the base varied, as in $ x^2 $ or $ x^3 $. Now swap the roles. In $ y = 2^x $, the values $ x = -2, -1, 0, 1, 2, 3 $ give $ y = \frac{1}{4}, \frac{1}{2}, 1, 2, 4, 8 $. Each step of 1 in $ x $ multiplies $ y $ by 2. Equal steps in $ x $ give equal ratios in $ y $, not equal differences. That is the defining feature.

Key termExponential function: a function $ f(x) = a^x $ with a constant base $ a > 0 $, $ a \neq 1 $; more generally $ f(x) = ka^x $.

The base must be positive so that $ a^x $ makes sense for every real $ x $ (what would $ (-4)^{1/2} $ be?), and it cannot be 1, because $ 1^x = 1 $ is just a horizontal line. Every curve $ y = a^x $ passes through $ (0, 1) $, because $ a^0 = 1 $. It is always positive, and it gets closer and closer to the line $ y = 0 $ without touching it: that line is the asymptote. Since $ \left(\frac{1}{2}\right)^x = 2^{-x} $, the decay curve is the reflection of $ y = 2^x $ in the $ y $-axis (Figure 1).

Exponential growth and decay

y = 2ˣy = (½)ˣy = 3ˣ

Figure 1. Every curve y = aˣ passes through (0, 1) and stays above its asymptote y = 0. A base above 1 gives growth, and a larger base climbs faster; y = (½)ˣ = 2⁻ˣ is the reflection of y = 2ˣ in the y-axis.

Transformations from Step 3 work as usual. For example, $ y = 3^x-9 $ is $ y = 3^x $ moved down 9, so its asymptote is $ y = -9 $. It crosses the $ y $-axis at $ (0, -8) $ and the $ x $-axis where $ 3^x = 9 $, at $ (2, 0) $.

Logarithms and their laws

You can solve $ 2^x = 32 $ by spotting that $ 32 = 2^5 $. But $ 2^x = 20 $ has no whole-number answer; it lies between 4 and 5. We need a name for “the power of 2 that gives 20”. That name is a logarithm.

Key termLogarithm: for $ a > 0 $, $ a \neq 1 $ and $ x > 0 $, $ \log_a x $ is the number $ y $ such that $ a^y = x $. Read it as “the power you raise $ a $ to, to get $ x $”.

So $ \log_3 81 = 4 $ because $ 3^4 = 81 $, $ \log_5 \frac{1}{25} = -2 $ because $ 5^{-2} = \frac{1}{25} $, and $ \log_8 2 = \frac{1}{3} $ because $ 8^{1/3} = 2 $. Each index law, read backwards, gives a fact about logs: $ a^0 = 1 $ means $ \log_a 1 = 0 $, and $ a^1 = a $ means $ \log_a a = 1 $. Since $ a^y $ is always positive, $ \log_a x $ exists only for $ x > 0 $.

Because a logarithm is an index, each index law becomes a log law. Write $ x = a^p $ and $ y = a^q $. Then $ xy = a^{p+q} $, $ x/y = a^{p-q} $ and $ x^n = a^{np} $. Reading off the powers gives the three laws.

\[ \begin{gathered} \log_a(xy) = \log_a x + \log_a y \\ \log_a\!\left(\frac{x}{y}\right) = \log_a x-\log_a y \\ \log_a(x^n) = n\log_a x \end{gathered} \]

For example, $ 3\ln 2 + \ln 5-\ln 4 = \ln 8 + \ln 5-\ln 4 = \ln\frac{8 \times 5}{4} = \ln 10 $. Move any multiplier inside as a power first, then add for products and subtract for quotients.

Worked example

Solve (a) $ \log_x 49 = 2 $ (b) $ \log_2 x + \log_2(x + 6) = 4 $.

Step 1(a) Index form: $ x^2 = 49 $, so $ x = \pm 7 $. A base must be positive, so $ x = 7 $.

Step 2(b) Combine into one log: $ \log_2\big(x(x + 6)\big) = 4 $, so $ x(x + 6) = 2^4 = 16 $.

Step 3$ x^2 + 6x-16 = 0 $, so $ (x + 8)(x-2) = 0 $ and $ x = 2 $ or $ x = -8 $.

Step 4Reject $ x = -8 $, because $ \log_2(-8) $ does not exist. Check $ x = 2 $: $ \log_2 2 + \log_2 8 = 1 + 3 = 4 $.

Always test your answers in the original equation. The algebra can produce a value that puts a negative number inside a logarithm.

Common mistakeInventing a law for $ \log(x + y) $. There is none: $ \log(x + y) $ is not $ \log x + \log y $, which equals $ \log(xy) $. In the same way $ (\log x)^2 $ is not $ 2\log x $, and $ \frac{\log x}{\log y} $ is not $ \log x-\log y $. The laws handle products, quotients and powers inside the log, nothing else.

Changing the base. Calculators offer only base 10 and base $ e $. To find $ \log_6 50 $, let $ y = \log_6 50 $, so $ 6^y = 50 $. Take $ \ln $ of both sides: $ y\ln 6 = \ln 50 $, so $ y = \frac{\ln 50}{\ln 6} = 2.183 $ to 4 significant figures. In general $ \log_a b = \frac{\ln b}{\ln a} $. IB lists this change-of-base rule; Cambridge 9709 does not require it, but the same “take $ \ln $ of both sides” move is all you need.

Exam tipCambridge writes $ \lg x $ for $ \log_{10} x $, which is the calculator’s log key; $ \ln x $ is base $ e $. Either base works for solving an equation, as long as you use the same one throughout.

Solving equations with the unknown in the index

When the unknown is in the index and the two sides are not powers of the same base, take logarithms of both sides. The power law brings the unknown down to the ground, and what is left is ordinary algebra. (This needs $ b > 0 $: $ a^x $ is never zero or negative.)

\[ a^x = b \;\Rightarrow\; x\ln a = \ln b \;\Rightarrow\; x = \frac{\ln b}{\ln a} \]

Worked example

Solve, giving answers to 3 significant figures, (a) $ 4^x = 30 $ (b) $ 3^{x+2} = 5^x $.

Step 1(a) Take $ \ln $: $ x\ln 4 = \ln 30 $, so $ x = \dfrac{\ln 30}{\ln 4} = 2.45 $.

Step 2(b) Take $ \ln $: $ (x + 2)\ln 3 = x\ln 5 $. The logs are just numbers, so this is a linear equation in $ x $.

Step 3Collect the $ x $ terms: $ 2\ln 3 = x\ln 5-x\ln 3 = x(\ln 5-\ln 3) $.

Step 4$ x = \dfrac{2\ln 3}{\ln 5-\ln 3} = \dfrac{2\ln 3}{\ln(5/3)} = 4.30 $.

Keep the logs exact until the last line. Rounding $ \ln 3 $ and $ \ln 5 $ early can cost the accuracy mark.

Some equations hide a quadratic. Since $ 2^{2x} = (2^x)^2 $, any equation built from $ 2^{2x} $ and $ 2^x $ is a quadratic in $ u = 2^x $, just as $ x^4-5x^2 + 4 = 0 $ is a quadratic in $ x^2 $.

Worked example

Solve $ 2^{2x}-2^{x+2}-12 = 0 $, giving your answer to 3 significant figures.

Step 1Let $ u = 2^x $. Then $ 2^{2x} = u^2 $ and $ 2^{x+2} = 2^x \times 2^2 = 4u $.

Step 2$ u^2-4u-12 = 0 $, so $ (u-6)(u + 2) = 0 $ and $ u = 6 $ or $ u = -2 $.

Step 3$ 2^x > 0 $ for every $ x $, so $ 2^x = -2 $ has no solution.

Step 4$ 2^x = 6 $ gives $ x = \dfrac{\ln 6}{\ln 2} = 2.58 $.

When you reject a negative value of $ u $, say why. “$ 2^x > 0 $ for all $ x $” is the reason examiners look for.

Common mistakeWriting $ 2^{x+2} $ as $ 2^x + 2 $ or $ 2^x + 4 $. The index law gives $ 2^{x+2} = 2^x \times 2^2 = 4 \times 2^x $. Adding in the index means multiplying the powers.

The number e and natural logarithms

Put \$1 in a bank paying 100% interest a year. Paid once, you end the year with \$2. Paid in two halves, $ 1.5^2 = \$2.25 $. Paid $ n $ times, you have $ \left(1 + \frac{1}{n}\right)^n $ dollars.

Payments per year, $ n $	1	2	4	12	365	1,000,000
$ \left(1 + \frac{1}{n}\right)^n $	2	2.25	2.4414	2.6130	2.7146	2.71828

More frequent payments help, but by less and less. The values settle towards a limit, $ e = 2.71828\ldots $, an irrational number like $ \pi $. What makes $ e $ the natural base for calculus is that the gradient of $ y = e^x $ at every point equals its height. You will prove this as $ \frac{d}{dx}(e^x) = e^x $ in the differentiation steps.

Key termNatural logarithm: $ \ln x = \log_e x $. So $ \ln x = y $ exactly when $ e^y = x $, and $ e^{\ln x} = x $ for $ x > 0 $, $ \ln(e^x) = x $, $ \ln 1 = 0 $ and $ \ln e = 1 $.

Since $ e^x $ and $ \ln x $ undo each other, they are inverse functions, and their graphs are reflections in the line $ y = x $ (Figure 2). Domain and range swap: $ e^x $ accepts any $ x $ and gives only positive outputs, while $ \ln x $ accepts only $ x > 0 $ and can give any real output.

y = eˣ and y = ln x are reflections in y = x

y = eˣy = ln xy = x

Figure 2. Reflecting in y = x swaps the coordinates: (0, 1) on y = eˣ becomes (1, 0) on y = ln x, and (1, e) becomes (e, 1). The asymptote y = 0 of eˣ becomes the asymptote x = 0 of ln x.

Worked example

Solve (a) $ \ln(3x-2) = 2 $ (b) $ e^{2x}-7e^x + 12 = 0 $. Give exact answers, then decimals to 3 significant figures.

Step 1(a) Undo $ \ln $ by raising $ e $ to each side: $ 3x-2 = e^2 $.

Step 2So $ x = \dfrac{e^2 + 2}{3} = 3.13 $. Check: $ 3x-2 = e^2 > 0 $, so the log exists.

Step 3(b) A quadratic in $ u = e^x $: $ u^2-7u + 12 = (u-3)(u-4) = 0 $.

Step 4$ e^x = 3 $ or $ e^x = 4 $, so $ x = \ln 3 = 1.10 $ or $ x = \ln 4 = 1.39 $.

The rule of thumb: apply whichever of $ e $ or $ \ln $ removes the outer function.

An economics example: continuous compounding. Invest \$2,000 at 6% a year for 8 years. Paid once a year, it grows to $ 2000 \times 1.06^8 = \$3{,}187.70 $. If the 6% is paid in $ n $ instalments a year and $ n $ grows without limit, the balance tends to $ A = Pe^{rt} = 2000e^{0.06 \times 8} = 2000e^{0.48} = \$3{,}232.15 $. That extra \$44.45 is the most that compounding more often can ever add. One year of continuous compounding multiplies by $ e^{0.06} = 1.0618 $, so the effective annual rate is about 6.18%, not 6%.

How long? Inequalities and growth models

“How many years until…?” questions want the smallest whole number that makes an inequality true. Take logs exactly as for an equation, with one extra rule. The log of any number between 0 and 1 is negative ($ \ln 0.85 = -0.1625\ldots $, because $ e^y < 1 $ only when $ y < 0 $). Dividing by a negative number reverses an inequality.

Worked example

A car bought for \$24,000 loses 15% of its value each year. After how many whole years is it first worth less than \$8,000?

Step 1After $ n $ years the value is $ 24000 \times 0.85^n $. We need $ 24000 \times 0.85^n < 8000 $, that is $ 0.85^n < \frac{1}{3} $.

Step 2Take $ \ln $ (it is increasing, so the sign stays): $ n\ln 0.85 < \ln\frac{1}{3} $.

Step 3Divide by $ \ln 0.85 $, which is negative, and reverse the sign: $ n > \dfrac{\ln(1/3)}{\ln 0.85} = 6.76\ldots $

Step 4Round up: $ n = 7 $. Check: $ 0.85^6 = 0.377 $, not below $ \frac{1}{3} $; $ 0.85^7 = 0.321 $, which is.

Figure 3 shows the value crossing the \$8,000 line between years 6 and 7.

A car losing 15% of its value each year value in $

V = 24,000 × 0.85ᵗ

Figure 3. The value falls by the same percentage every year, so the curve flattens but never reaches 0. It first drops below $8,000 in year 7, matching n > 6.76.

In science and economics, growth is often written in base $ e $: $ y = Ae^{kt} $, where $ A $ is the value at $ t = 0 $ and $ k > 0 $ means growth, $ k < 0 $ decay. Doubling means $ Ae^{kT} = 2A $. The $ A $ cancels, leaving $ e^{kT} = 2 $, so $ T = \frac{\ln 2}{k} $. The starting value has vanished, so the doubling time is the same whenever you start counting. That is the signature of exponential change. A half-life works the same way with $ e^{kT} = \frac{1}{2} $, giving $ T = \frac{\ln 2}{|k|} $.

\[ y = Ae^{kt} \qquad \text{doubling time or half-life} = \frac{\ln 2}{|k|} \]

An economics example: the Rule of 70. An economy growing continuously at rate $ r $ doubles in $ \frac{\ln 2}{r} $ years. Since $ \ln 2 = 0.693\ldots $, growth of $ g $ per cent a year doubles output in about $ \frac{69.3}{g} \approx \frac{70}{g} $ years. At 2.5% a year that is $ 70 \div 2.5 = 28 $ years. With growth paid once a year, the exact answer is $ \frac{\ln 2}{\ln 1.025} = 28.07 $ years, so using 70 rather than 69.3 roughly allows for growth paid once a year.

Turning curves into straight lines

If you suspect that data follow $ y = kx^n $ or $ y = ka^x $, a plot of $ y $ against $ x $ is a curve, and constants are hard to read from a curve. Taking logs turns either model into a straight line $ Y = mX + c $ (Step 4), whose gradient and intercept you can read.

Model	Take $ \ln $	Plot	Gradient, intercept
$ y = kx^n $	$ \ln y = n\ln x + \ln k $	$ \ln y $ against $ \ln x $	$ n $, $ \ln k $
$ y = ka^x $	$ \ln y = (\ln a)x + \ln k $	$ \ln y $ against $ x $	$ \ln a $, $ \ln k $

Worked example

Variables satisfy $ y = ka^x $. The graph of $ \ln y $ against $ x $ is a straight line through $ (1, 2.3) $ and $ (5, 4.7) $. Find $ a $ and $ k $ to 3 significant figures.

Step 1The gradient is $ \ln a $: $ \ln a = \dfrac{4.7-2.3}{5-1} = 0.6 $, so $ a = e^{0.6} = 1.82 $.

Step 2Substitute a point into $ \ln y = 0.6x + \ln k $: $ 2.3 = 0.6 + \ln k $, so $ \ln k = 1.7 $.

Step 3$ k = e^{1.7} = 5.47 $. The model is $ y = 5.47 \times 1.82^x $.

The intercept is $ \ln k $, not $ k $. Finish with $ k = e^{\text{intercept}} $, or $ 10^{\text{intercept}} $ if the axis is $ \lg y $.

Economists use the same trick. On a log–log plot, the demand curve $ Q = kP^n $ becomes the straight line $ \ln Q = \ln k + n\ln P $. Its gradient $ n $ is the price elasticity of demand, the same at every price: if $ n = -1.5 $, a 1% price rise cuts quantity demanded by about 1.5%. In the same way, a GDP series plotted on a log scale is a straight line when the growth rate is constant.

Exam tipCambridge 9709 tests logarithms and exponentials on Papers 2 and 3, not Paper 1, and linearising data is a favourite. IB covers log laws and change of base in topic 1 and exponential models in topic 2. AP Calculus assumes all of it from Unit 2 onwards, when you differentiate $ e^x $ and $ \ln x $.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Evaluate (a) $ \log_2 64 $ (b) $ \lg 0.01 $ (c) $ \log_{27} 3 $. Show worked answer

(a) $ 2^6 = 64 $, so $ \log_2 64 = 6 $.
(b) $ 10^{-2} = 0.01 $, so $ \lg 0.01 = -2 $.
(c) $ 27^{1/3} = 3 $, so $ \log_{27} 3 = \frac{1}{3} $.

EasyQ2. Write $ 2\lg 5 + \lg 4-\lg 10 $ as a single logarithm and evaluate it. Show worked answer

Power law first: $ 2\lg 5 = \lg 25 $.
Combine: $ \lg\dfrac{25 \times 4}{10} = \lg 10 $.
$ \lg 10 = 1 $.

MediumQ3. Solve $ 5^{2x+1} = 40 $, giving your answer to 3 significant figures. Show worked answer

Take $ \ln $: $ (2x + 1)\ln 5 = \ln 40 $.
$ 2x + 1 = \dfrac{\ln 40}{\ln 5} = 2.2920\ldots $
$ x = \dfrac{2.2920\ldots-1}{2} = 0.646 $.

MediumQ4. Solve $ \lg x + \lg(x-3) = 1 $. Show worked answer

Combine: $ \lg\big(x(x-3)\big) = 1 $, so $ x(x-3) = 10^1 = 10 $.
$ x^2-3x-10 = 0 $, so $ (x-5)(x + 2) = 0 $.
$ x = -2 $ would need $ \lg(-2) $, so reject it. The solution is $ x = 5 $.
Check: $ \lg 5 + \lg 2 = \lg 10 = 1 $.

MediumQ5. Solve $ e^{2x}-5e^x-14 = 0 $, giving an exact answer and a decimal to 3 significant figures. Show worked answer

Let $ u = e^x $: $ u^2-5u-14 = 0 $, so $ (u-7)(u + 2) = 0 $.
$ e^x = -2 $ is impossible, because $ e^x > 0 $ for all $ x $.
$ e^x = 7 $ gives $ x = \ln 7 = 1.95 $.

ExamQ6. The number of users of an app, $ P $, is modelled by $ P = 400e^{kt} $, where $ t $ is the time in months. When $ t = 3 $, $ P = 600 $. (a) Find the exact value of $ k $, and its value to 3 significant figures. [2] (b) Find the time it takes the number of users to double. [2] (c) Find the first whole number of months after which $ P $ exceeds 5000. [3] (d) State what straight-line graph the model gives, with its gradient and intercept. [2] Show worked answer

(a) $ 600 = 400e^{3k} $, so $ e^{3k} = 1.5 $ [M1] and $ k = \frac{1}{3}\ln 1.5 = 0.135 $ [A1].
(b) Doubling time $ = \dfrac{\ln 2}{k} $ [M1] $ = \dfrac{3\ln 2}{\ln 1.5} = 5.13 $ months [A1].
(c) $ 400e^{kt} > 5000 $, so $ e^{kt} > 12.5 $ [M1]. $ t > \dfrac{\ln 12.5}{k} = \dfrac{3\ln 12.5}{\ln 1.5} = 18.69\ldots $ [A1]. So after 19 months [A1].
(d) $ \ln P = \ln 400 + kt $: plotting $ \ln P $ against $ t $ gives a straight line [B1] with gradient $ k = 0.135 $ and intercept $ \ln 400 = 5.99 $ [B1].
Total: 9 marks.

Check yourself

1. Which expression is equal to $ \log_a(xy) $?

A log is an index, and multiplying numbers adds their indices: $ a^p \times a^q = a^{p+q} $.

2. You reach $ n\ln 0.7 < \ln 0.2 $. What comes next?

$ \ln 0.7 $ is negative, so dividing by it reverses the inequality. That makes sense: shrinking by 0.7 each time, you need more steps to fall below 0.2.

3. For the model $ y = ka^x $, you plot $ \ln y $ against $ x $. What is the gradient of the line?

Taking logs gives $ \ln y = (\ln a)x + \ln k $, so the gradient is $ \ln a $ and the intercept is $ \ln k $.

Your score: 0 out of 3

The short version

$ y = a^x $ with $ a > 0 $, $ a \neq 1 $ passes through $ (0, 1) $, stays positive and has asymptote $ y = 0 $. A base above 1 is growth; between 0 and 1 is decay.
$ \log_a x = y $ means $ a^y = x $. Logs exist only for $ x > 0 $. Products, quotients and powers inside a log become sums, differences and multiples; there is no law for $ \log(x + y) $.
Take logs of both sides to solve $ a^x = b $. Equations in $ a^{2x} $ and $ a^x $ are quadratics in $ u = a^x $; reject $ u \le 0 $.
$ e \approx 2.718 $, $ \ln x = \log_e x $, and $ e^x $ and $ \ln x $ are inverses. $ y = Ae^{kt} $ doubles or halves every $ \frac{\ln 2}{|k|} $, and dividing by the log of a number below 1 flips an inequality.
$ y = kx^n $: plot $ \ln y $ against $ \ln x $. $ y = ka^x $: plot $ \ln y $ against $ x $. The intercept is $ \ln k $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

Model	Take \( \ln \)	Plot	Gradient, intercept
\( y = kx^n \)	\( \ln y = n\ln x + \ln k \)	\( \ln y \) against \( \ln x \)	\( n \), \( \ln k \)
\( y = ka^x \)	\( \ln y = (\ln a)x + \ln k \)	\( \ln y \) against \( x \)	\( \ln a \), \( \ln k \)

Step 5: Exponentials and Logarithms

Exponential functions and their graphs

Logarithms and their laws

Solving equations with the unknown in the index

The number e and natural logarithms

How long? Inequalities and growth models

Turning curves into straight lines

Practise: 6 questions

Check yourself

The short version

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