Logarithms — The Basics
Definition · Index ↔ log form · Common log · Special values · §5.1
The fundamental definition
A logarithm answers the question: "What power do I raise the base to, in order to get this number?"
\(\log_b a = x \iff b^x = a\)
Conditions: \(b > 0\), \(b \neq 1\), and \(a > 0\). You can only take the log of a positive number.
Read it as: "log base b of a equals x" means "b to the power x equals a".
Index form
\(b^x = a\)
⟷
convert
Log form
\(\log_b a = x\)
Index: \(2^3 = 8\)
→ Log: \(\log_2 8 = 3\)
Index: \(10^2 = 100\)
→ Log: \(\log_{10} 100 = 2\)
Log: \(\log_3 81 = 4\)
→ Index: \(3^4 = 81\)
Special values — memorise these
| Expression | Value | Why? |
|---|---|---|
| \(\log_b b\) | 1 | because \(b^1 = b\) |
| \(\log_b 1\) | 0 | because \(b^0 = 1\) |
| \(\log_b b^n\) | n | because \(b^n = b^n\) |
| \(b^{\log_b a}\) | a | inverse relationship |
Quick checks: \(\log_{10} 1000 = 3\) since \(10^3 = 1000\). \(\log_5 \frac{1}{25} = -2\) since \(5^{-2} = \frac{1}{25}\).
Common logarithm (base 10)
When no base is written, the convention in IGCSE is base 10. Written as lg or log.
\(\lg x = \log_{10} x\)
Key values of \(\lg\):
\(\lg 1 = 0\) · \(\lg 10 = 1\) · \(\lg 100 = 2\)
\(\lg 0.1 = -1\) · \(\lg 0.01 = -2\)
\(\lg 1 = 0\) · \(\lg 10 = 1\) · \(\lg 100 = 2\)
\(\lg 0.1 = -1\) · \(\lg 0.01 = -2\)
Domain: \(\lg x\) is only defined for \(x > 0\). Never take a log of zero or a negative number.
Interactive — Index ↔ Log Converter
For \(b^x = a\), adjust the sliders and see both forms
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Evaluate — test yourself
Laws of Logarithms
Product · Quotient · Power · Change of base · §5.2
Product Law
\(\log_b(mn) = \log_b m + \log_b n\)
Log of a product = sum of logs. Follows from \(b^p \cdot b^q = b^{p+q}\).
\(\log_2 8 + \log_2 4 = \log_2 32 = 5\)
Quotient Law
\(\log_b\!\left(\dfrac{m}{n}\right) = \log_b m - \log_b n\)
Log of a quotient = difference of logs. Follows from \(b^p \div b^q = b^{p-q}\).
\(\log_3 27 - \log_3 3 = \log_3 9 = 2\)
Power Law
\(\log_b(m^n) = n\log_b m\)
Log of a power = power × log. This is the most used law for solving equations.
\(\log_2 8^3 = 3\log_2 8 = 3 \times 3 = 9\)
Change of Base
Convert a logarithm from any base to a computable base (usually base 10 or e).
\(\log_b a = \dfrac{\log_c a}{\log_c b} = \dfrac{\lg a}{\lg b} = \dfrac{\ln a}{\ln b}\)
Evaluate \(\log_3 7\)
Apply
\(\log_3 7 = \dfrac{\lg 7}{\lg 3}\)
Calc
\(= \dfrac{0.8451}{0.4771}\)
✓
\(\approx 1.771\)
Common errors — what NOT to write
✗ \(\log(m + n) \neq \log m + \log n\)
✗ \(\log(m - n) \neq \log m - \log n\)
✗ \(\dfrac{\log m}{\log n} \neq \log\!\left(\dfrac{m}{n}\right)\)
✗ \((\log m)(\log n) \neq \log(mn)\)
Simplify \(\log_2 48 - \log_2 3\)
Quotient
\(= \log_2\!\left(\dfrac{48}{3}\right) = \log_2 16\)
✓
\(= \log_2 2^4 = 4\)
Combining multiple laws — worked example
Write \(2\lg a + \lg b - \lg c\) as a single logarithm
Power
\(2\lg a = \lg a^2\)
Product
\(\lg a^2 + \lg b = \lg(a^2 b)\)
Quotient
\(\lg(a^2 b) - \lg c = \lg\!\left(\dfrac{a^2 b}{c}\right)\)
✓
\(\lg\!\left(\dfrac{a^2 b}{c}\right)\)
Given \(\lg 2 = p\) and \(\lg 3 = q\), express \(\lg 72\) in terms of p and q
Factorise
\(72 = 8 \times 9 = 2^3 \times 3^2\)
Product
\(\lg 72 = \lg 2^3 + \lg 3^2\)
Power
\(= 3\lg 2 + 2\lg 3\)
✓
\(= 3p + 2q\)
Natural Logarithms & the Number e
e ≈ 2.718 · ln x = log_e x · inverse relationship · §5.3
The number e
Euler's number e ≈ 2.71828… is an irrational constant that arises naturally in calculus, compound growth, and many other contexts. Like π, it is a fundamental mathematical constant.
\(e = \lim_{n\to\infty}\!\left(1 + \tfrac{1}{n}\right)^n \approx 2.71828\)
The natural logarithm uses base e and is written ln:
\(\ln x = \log_e x\)
Key property: \(y = e^x\) and \(y = \ln x\) are inverse functions. Their graphs are reflections of each other in the line \(y = x\).
Essential identities
\(\ln e = 1\)since \(e^1 = e\)
\(\ln 1 = 0\)since \(e^0 = 1\)
\(\ln e^k = k\)power law
\(e^{\ln x} = x\)inverse cancellation
Simplify \(e^{3\ln 2}\)
Power law
\(e^{3\ln 2} = e^{\ln 2^3} = e^{\ln 8}\)
✓
Since \(e^{\ln x} = x\): \(= 8\)
Solve \(e^{2x} = 15\)
ln both sides
\(\ln e^{2x} = \ln 15\)
Simplify
\(2x = \ln 15\)
✓
\(x = \dfrac{\ln 15}{2} \approx 1.354\)
Solve \(\ln(3x - 1) = 4\)
Exponentiate
Raise e to both sides: \(e^{\ln(3x-1)} = e^4\)
Simplify
\(3x - 1 = e^4\)
✓
\(x = \dfrac{e^4 + 1}{3} \approx 18.53\)
All three log laws also apply to ln
Product
\(\ln(mn) = \ln m + \ln n\)
Quotient
\(\ln\!\left(\dfrac{m}{n}\right) = \ln m - \ln n\)
Power
\(\ln(m^n) = n\ln m\)
Graphs of Log & Exponential Functions
Key features · Transformations · Asymptotes · §5.4
y = aˣ — Exponential graph
y-intercept(0, 1) always — since \(a^0 = 1\)
asymptote\(y = 0\) (the x-axis) as \(x \to -\infty\)
a > 1Increasing — rises steeply to the right
0 < a < 1Decreasing — falls steeply to the right
range\(y > 0\) for all x
y = log_a x — Logarithmic graph
x-intercept(1, 0) always — since \(\log_a 1 = 0\)
asymptote\(x = 0\) (the y-axis) as \(x \to 0^+\)
a > 1Increasing — slow growth
domain\(x > 0\) only
inverseReflection of \(y = a^x\) in \(y = x\)
Interactive — Graph Explorer
Explore \(y = a \cdot b^{x+c} + d\) and its transformations
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Transformations applied to log & exponential graphs
y = f(x) + k
Translation by \(\binom{0}{k}\). Shifts graph up/down. Asymptote moves from \(y=0\) to \(y=k\) (for exp) or \(x=0\) unchanged (for log).
y = f(x + k)
Translation by \(\binom{-k}{0}\). Shifts graph left/right. Asymptote moves: \(x = 0\) becomes \(x = -k\) for log graph.
y = af(x)
Stretch by factor a parallel to y-axis. For exp: y-intercept moves from 1 to a. Asymptote unchanged.
y = −f(x)
Reflection in the x-axis. Exp graph flips below x-axis with asymptote \(y = 0\) still. Log: x-intercept stays at (1, 0), but graph flips.
Solving Log & Exponential Equations
Three strategies · Step-by-step methods · §5.5
Strategy 1
Take logs of both sides
For \(a^x = b\) — take log (base 10 or e) of both sides, then use the power law to bring x down.
\(a^x = b \Rightarrow x\lg a = \lg b \Rightarrow x = \dfrac{\lg b}{\lg a}\)
Strategy 2
Convert to exponential form
For \(\log_b(f(x)) = k\) — rewrite as \(f(x) = b^k\) (no log needed), then solve for x.
\(\log_2(x-1) = 3 \Rightarrow x-1 = 8 \Rightarrow x = 9\)
Strategy 3
Collect logs, then convert
For equations with multiple log terms — use the laws to combine into a single log, then convert to exponential form.
\(\log x + \log 3 = 2 \Rightarrow \log(3x) = 2 \Rightarrow 3x = 100\)
Solve \(3^x = 20\) (Strategy 1)
Take lg
\(\lg 3^x = \lg 20\)
Power law
\(x \lg 3 = \lg 20\)
✓
\(x = \dfrac{\lg 20}{\lg 3} \approx 2.727\)
Solve \(2^{x+1} = 5^{x-1}\)
Take lg
\((x+1)\lg 2 = (x-1)\lg 5\)
Expand
\(x\lg 2 + \lg 2 = x\lg 5 - \lg 5\)
Collect x
\(x(\lg 2 - \lg 5) = -\lg 5 - \lg 2\)
✓
\(x = \dfrac{-(\lg 5 + \lg 2)}{\lg 2 - \lg 5} = \dfrac{-\lg 10}{\lg(2/5)} = \dfrac{-1}{\lg 0.4} \approx 2.513\)
Solve \(\lg(x+6) + \lg(x-3) = 1\) (Strategy 3)
Combine
\(\lg[(x+6)(x-3)] = 1\)
Convert
\((x+6)(x-3) = 10^1 = 10\)
Expand
\(x^2 + 3x - 18 = 10\)
Rearrange
\(x^2 + 3x - 28 = 0\)
Factorise
\((x+7)(x-4) = 0 \Rightarrow x=-7\) or \(x=4\)
Check
\(x = -7\): \(\lg(-1)\) undefined ✗ Reject \(x = -7\)
✓
\(x = 4\)
Simultaneous equations with logs
Solve: \(\lg x + \lg y = 3\) and \(\lg x - \lg y = 1\)
Sum
Add: \(2\lg x = 4 \Rightarrow \lg x = 2 \Rightarrow x = 100\)
Diff
Subtract: \(2\lg y = 2 \Rightarrow \lg y = 1 \Rightarrow y = 10\)
✓
\(x = 100,\quad y = 10\)
Critical: always check for invalid solutions. After solving any equation involving logs, substitute back. If any log argument becomes ≤ 0, reject that solution. This catches extraneous roots introduced when combining logs or using the quadratic formula.
Reducing to Linear Form
y = ax^n and y = Ab^x · log graphs · finding constants · §5.6
The big idea
Relationships like \(y = ax^n\) and \(y = Ab^x\) are not linear, so a scatter plot of y against x is a curve. By taking logs, we convert them to straight-line form \(Y = mX + c\), which allows us to find the constants graphically.
Type 1 — Power law: y = ax^n
Take lg
\(\lg y = \lg(ax^n)\)
Expand
\(\lg y = \lg a + n\lg x\)
Linear
Plot \(\lg y\) against \(\lg x\)
Gradient = \(n\) · y-intercept = \(\lg a\)
Gradient = \(n\) · y-intercept = \(\lg a\)
Type 2 — Exponential: y = Ab^x
Take lg
\(\lg y = \lg(Ab^x)\)
Expand
\(\lg y = x\lg b + \lg A\)
Linear
Plot \(\lg y\) against \(x\)
Gradient = \(\lg b\) · y-intercept = \(\lg A\)
Gradient = \(\lg b\) · y-intercept = \(\lg A\)
Summary table — what to plot and what to read off
| Equation | Plot | Gradient | Y-intercept |
|---|---|---|---|
| \(y = ax^n\) | \(\lg y\) vs \(\lg x\) | \(n\) | \(\lg a\) |
| \(y = Ab^x\) | \(\lg y\) vs \(x\) | \(\lg b\) | \(\lg A\) |
| \(y = Ae^{bx}\) | \(\ln y\) vs \(x\) | \(b\) | \(\ln A\) |
| \(y = a + bx^2\) | \(y\) vs \(x^2\) | \(b\) | \(a\) |
Reading a and b from a graph: Once you have gradient m and y-intercept c from the straight line, reverse the substitution: e.g. for \(y=ax^n\), \(n = m\) and \(a = 10^c\).
The graph of \(\lg y\) against \(\lg x\) is a straight line with gradient 3 and \(\lg y\)-intercept \(-1\). Find y in terms of x.
Line form
\(\lg y = 3\lg x + (-1)\)
Power law
\(\lg y = \lg x^3 + \lg 10^{-1}\)
Combine
\(\lg y = \lg\!\left(\dfrac{x^3}{10}\right)\)
✓
\(y = \dfrac{x^3}{10} = 0.1x^3\)
i.e. \(a = 0.1\) and \(n = 3\) in \(y = ax^n\).
Interactive — Linear Law Visualiser
Plot \(\lg y\) vs \(\lg x\) for \(y = ax^n\) — see how the straight line encodes a and n
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Practice Questions
3 Easy · 3 Medium · 3 Hard — Logs & Exponentials
Easy 1
[2] Simplify \(\log_2 8 + \log_2 4\).
SOLUTION
Law
\(\log_2 8 + \log_2 4 = \log_2(8\times4)=\log_2 32 = \boldsymbol{5}\)
Easy 2
[2] Solve \(3^x = 20\), giving your answer to 3 significant figures.
SOLUTION
Take log
\(x\ln3=\ln20 \Rightarrow x=\dfrac{\ln20}{\ln3}=\boldsymbol{2.73}\) (3 s.f.)
Easy 3
[2] Write \(\log\dfrac{x^3}{y^2}\) in terms of \(\log x\) and \(\log y\).
SOLUTION
Laws
\(\log(x^3)-\log(y^2) = \boldsymbol{3\log x - 2\log y}\)
Medium 1
[4] Solve \(2\log x - \log(x+6) = 1\), where logarithms are base 10.
SOLUTION
Rewrite
\(\log\!\left(\dfrac{x^2}{x+6}\right)=1 \Rightarrow \dfrac{x^2}{x+6}=10\)
Solve
\(x^2=10x+60 \Rightarrow x^2-10x-60=0\)
\(x=\dfrac{10\pm\sqrt{340}}{2}=5\pm\sqrt{85}\)
\(x=\dfrac{10\pm\sqrt{340}}{2}=5\pm\sqrt{85}\)
Answer
Reject negative (\(\log\) undefined): \(\boldsymbol{x=5+\sqrt{85}\approx14.2}\)
Medium 2
[4] Solve \(e^{2x}-5e^x+6=0\).
SOLUTION — substitution
Let \(u=e^x\)
\(u^2-5u+6=0 \Rightarrow (u-2)(u-3)=0 \Rightarrow u=2\) or \(u=3\)
Answer
\(e^x=2 \Rightarrow x=\ln2\) or \(e^x=3 \Rightarrow x=\ln3\)
\(\boldsymbol{x=\ln2}\) or \(\boldsymbol{x=\ln3}\)
\(\boldsymbol{x=\ln2}\) or \(\boldsymbol{x=\ln3}\)
Medium 3
[4] Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express \(\log_a 54\) in terms of \(p\) and \(q\).
SOLUTION
Factorise
\(54=2\times27=2\times3^3\)
Log laws
\(\log_a54=\log_a2+3\log_a3=\boldsymbol{p+3q}\)
Hard 1
[5] Solve \(\log_3(2x+1)-\log_9 x=1\).
SOLUTION — change of base
Same base
\(\log_9x=\dfrac{\log_3x}{\log_3 9}=\dfrac{\log_3x}{2}\)
Substitute
\(\log_3(2x+1)-\dfrac{\log_3x}{2}=1\)
\(\log_3\!\left(\dfrac{2x+1}{\sqrt{x}}\right)=1 \Rightarrow \dfrac{2x+1}{\sqrt{x}}=3\)
\(\log_3\!\left(\dfrac{2x+1}{\sqrt{x}}\right)=1 \Rightarrow \dfrac{2x+1}{\sqrt{x}}=3\)
Solve
\(2x+1=3\sqrt{x}\). Let \(u=\sqrt{x}\): \(2u^2-3u+1=0 \Rightarrow (2u-1)(u-1)=0\)
\(\sqrt{x}=\tfrac12\Rightarrow x=\tfrac14\) or \(\sqrt{x}=1\Rightarrow x=1\)
\(\boldsymbol{x=\tfrac{1}{4}}\) or \(\boldsymbol{x=1}\)
\(\sqrt{x}=\tfrac12\Rightarrow x=\tfrac14\) or \(\sqrt{x}=1\Rightarrow x=1\)
\(\boldsymbol{x=\tfrac{1}{4}}\) or \(\boldsymbol{x=1}\)
Hard 2
[5] The variables \(x\) and \(y\) satisfy \(y=Ab^x\). A straight line graph of \(\lg y\) against \(x\) passes through \((0,\,1.2)\) and \((4,\,3.6)\). Find \(A\) and \(b\).
SOLUTION — linear graph of logs
Linearise
\(\lg y=\lg A+x\lg b\). This is \(Y=mX+c\) where \(Y=\lg y,\;X=x\).
From graph
y-intercept \(c=\lg A=1.2 \Rightarrow A=10^{1.2}\approx\boldsymbol{15.8}\)
Gradient \(=\lg b=\dfrac{3.6-1.2}{4}=0.6 \Rightarrow b=10^{0.6}\approx\boldsymbol{3.98}\)
Gradient \(=\lg b=\dfrac{3.6-1.2}{4}=0.6 \Rightarrow b=10^{0.6}\approx\boldsymbol{3.98}\)
Hard 3
[5] Solve simultaneously: \(\log(x+y)=1\) and \(\log x+\log(2y-1)=\log 12\).
SOLUTION
Eq 1
\(x+y=10\) ...[1]
Eq 2
\(x(2y-1)=12\) ...[2]
Solve
From [1]: \(x=10-y\). Sub into [2]: \((10-y)(2y-1)=12\)
\(20y-10-2y^2+y=12 \Rightarrow 2y^2-21y+22=0 \Rightarrow (2y-11)(y-2)=0\)
\(y=\tfrac{11}{2}\Rightarrow x=\tfrac{9}{2}\) or \(y=2\Rightarrow x=8\)
\(20y-10-2y^2+y=12 \Rightarrow 2y^2-21y+22=0 \Rightarrow (2y-11)(y-2)=0\)
\(y=\tfrac{11}{2}\Rightarrow x=\tfrac{9}{2}\) or \(y=2\Rightarrow x=8\)
Check
Both valid (all log arguments positive). \(\boldsymbol{x=\tfrac92,y=\tfrac{11}{2}}\) or \(\boldsymbol{x=8,y=2}\)
Past Year Paper Questions
Cambridge 0606 — Logarithms
0606
[11]
(a) Solve \(4^x-6\cdot2^x+8=0\). [4]
(b) Given \(\lg y=3\lg x-\lg 5+2\), express \(y\) in terms of \(x\). [3]
(c) Solve \(\log_4(3x-2)=2-\log_4 x\). [4]
(b) Given \(\lg y=3\lg x-\lg 5+2\), express \(y\) in terms of \(x\). [3]
(c) Solve \(\log_4(3x-2)=2-\log_4 x\). [4]
FULL SOLUTION
(a)
Let \(u=2^x\): \(u^2-6u+8=0\Rightarrow(u-2)(u-4)=0\)
\(2^x=2\Rightarrow x=1\) or \(2^x=4\Rightarrow x=2\). \(\boldsymbol{x=1}\) or \(\boldsymbol{x=2}\)
\(2^x=2\Rightarrow x=1\) or \(2^x=4\Rightarrow x=2\). \(\boldsymbol{x=1}\) or \(\boldsymbol{x=2}\)
(b)
\(\lg y=\lg x^3-\lg5+2=\lg\!\left(\dfrac{x^3}{5}\right)+\lg100\)
\(y=\dfrac{100x^3}{5}=\boldsymbol{20x^3}\)
\(y=\dfrac{100x^3}{5}=\boldsymbol{20x^3}\)
(c)
\(\log_4(3x-2)+\log_4x=2 \Rightarrow \log_4(x(3x-2))=2\)
\(3x^2-2x=16 \Rightarrow 3x^2-2x-16=0\Rightarrow(3x+8)(x-2)=0\)
\(x=2\) (reject \(x=-\tfrac83\) as log undefined). \(\boldsymbol{x=2}\)
\(3x^2-2x=16 \Rightarrow 3x^2-2x-16=0\Rightarrow(3x+8)(x-2)=0\)
\(x=2\) (reject \(x=-\tfrac83\) as log undefined). \(\boldsymbol{x=2}\)
0606 Style
Change of Base & Log Laws
[8]
(a) Show that \(\log_a b = \dfrac{1}{\log_b a}\). [2]
(b) Solve \(\log_3 x + \log_x 27 = 4\). [4]
(c) Without a calculator, evaluate \(\log_8 16\). [2]
(b) Solve \(\log_3 x + \log_x 27 = 4\). [4]
(c) Without a calculator, evaluate \(\log_8 16\). [2]
FULL WORKED SOLUTION
(a)
\(\log_a b = \frac{\ln b}{\ln a}\) and \(\log_b a = \frac{\ln a}{\ln b}\). Product: \(\frac{\ln b}{\ln a}\cdot\frac{\ln a}{\ln b}=1\Rightarrow\log_a b=\frac{1}{\log_b a}\) ✓
(b)
Let \(t=\log_3 x\). Then \(\log_x 27=\frac{3}{t}\) (by change of base).
\(t+\frac{3}{t}=4\Rightarrow t^2-4t+3=0\Rightarrow(t-1)(t-3)=0\)
\(t=1\Rightarrow x=3\); \(t=3\Rightarrow x=27\).
\(\boldsymbol{x=3}\) or \(\boldsymbol{x=27}\)
\(t+\frac{3}{t}=4\Rightarrow t^2-4t+3=0\Rightarrow(t-1)(t-3)=0\)
\(t=1\Rightarrow x=3\); \(t=3\Rightarrow x=27\).
\(\boldsymbol{x=3}\) or \(\boldsymbol{x=27}\)
(c)
\(\log_8 16=\frac{\log_2 16}{\log_2 8}=\frac{4}{3}\). So \(\boldsymbol{\log_8 16=\frac{4}{3}}\)
0606 Style
Exponential Equation with Substitution
[6]
By using the substitution \(y = 2^x\), or otherwise, solve \(2^{2x} - 6(2^x) + 8 = 0\). [6]
FULL WORKED SOLUTION
Substitute
Let \(y=2^x\). Then \(2^{2x}=y^2\).
\(y^2-6y+8=0\Rightarrow(y-2)(y-4)=0\)
\(y^2-6y+8=0\Rightarrow(y-2)(y-4)=0\)
Solve
\(y=2\Rightarrow 2^x=2\Rightarrow\boldsymbol{x=1}\)
\(y=4\Rightarrow 2^x=4\Rightarrow\boldsymbol{x=2}\)
\(y=4\Rightarrow 2^x=4\Rightarrow\boldsymbol{x=2}\)