🧮 Logarithmic Functions
Logarithms compress vast ranges — from sound intensity to earthquake energy — into numbers engineers can actually work with, and unlock any equation with the unknown trapped in an exponent.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
🧮Laws of LogsMultiply → add, etc.
🔍Solving EquationsUnknown inside or as a power
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Apply the three laws of logarithms to combine and simplify expressions; solve equations where the unknown appears as an exponent or inside a log; and use log scales (dB, pH, Richter) that compress enormous ranges into workable numbers.
💡 A logarithm answers one question: "what power do I raise the base to, to get this number?" Every log law follows from that single idea.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
🧮
Laws of Logarithms
Three laws turn multiplication into addition, division into subtraction, and powers into multipliers.
🔧Log Combiner — simplify any expressionEnter A, B and n, then press Combine
Enter values, choose a law, then press Combine to see it expanded and evaluated…
📖 Key Vocabulary
Logarithm — the power you raise the base to, to get a given number.
Base — the number being raised to a power (10 for log, e for ln).
Change of base — logbA = ln A ÷ ln b, needed for bases your calculator lacks.
log 1 = 0 — any base raised to the power 0 gives 1, for any base.
👁️ Visual aid: every log law trades a harder operation (×, ÷, powers) for an easier one (+, −, ×constant).
✏️Worked Example 1 — Combining logs (multiplication & division)Type: Combine
Question Simplify log 8 + log 5 − log 2 as a single logarithm.
Apply the addition law
log8 + log5 = log(8×5) = log40 — two added logs of the same base become the log of a product, so deal with the "+" first.Apply the subtraction law
log40 − log2 = log(40/2) — the remaining "−" becomes a division inside a single log.Simplify inside the log
40 ÷ 2 = 20 — do the arithmetic first so the log argument is a single tidy number.Answer = log 20 — one logarithm, one number inside it, nothing left to combine.
🔢Worked Example 2 — Using the power lawType: Power law
Question Express 3log2 + log5 as a single logarithm.
Apply the power law first
3log2 = log(2³) = log8 — a number multiplying a log always moves inside as an exponent before anything else is combined; 2³ = 8 confirms the value.Apply the addition law
log8 + log5 = log(8×5) — now that both terms are plain logs (coefficient 1), the addition law turns the sum into the log of a product.Simplify inside the log
8 × 5 = 40 — multiply out the argument to leave a single number.Answer = log 40 — always clear the power law before applying addition or subtraction, or the laws won't combine correctly.
🔁Worked Example 3 — Change of baseType: Change of base
Question Evaluate log₅(30) using natural logs.
Identify the problem A calculator has no "log base 5" button — only log₁₀ and ln — so the base needs changing to one the calculator understands.
Apply the change of base formula
log₅30 = ln30 ÷ ln5 — any base b can be rewritten this way: log_b(x) = ln x ÷ ln b. Either ln or log₁₀ works, as long as both top and bottom use the same one.Evaluate each natural log
ln30 ≈ 3.4012, ln5 ≈ 1.6094 — find each value separately before dividing.Divide
3.4012 ÷ 1.6094Answer ≈ 2.113 — useful whenever a calculator only has log₁₀ and ln buttons.
⚙️Worked Example 4 — Engineering context (sound intensity, dB)Type: Engineering
Question Sound level is L = 10log₁₀(I/I₀). Find I/I₀ for a jet engine at L = 140 dB.
Isolate the log term
log₁₀(I/I₀) = 140 ÷ 10 = 14 — divide both sides by 10 to undo the multiplier in front of the log, leaving the log on its own.Convert from log form to exponential form
log₁₀(I/I₀) = 14 → I/I₀ = 10¹⁴ — log_b(x) = y means x = bʸ, so raise the base (10) to the power on the right.Answer The jet's intensity is 10¹⁴ times the reference intensity — logs compress this vast range into a manageable 140 dB.
🧠 Common slip: writing log(A+B) = logA + logB. This is never true — the addition law only applies to log(AB), a product, not a sum.
🔍
Solving Logarithmic Equations
Two flavours: unknown as an exponent (take logs), or unknown inside a log (convert to exponential form).
🔧Log Equation SolverSolve a·bˣ = c for x
Enter a, b and c for a·bˣ = c, then press Solve to see the full log working…
📖 Key Vocabulary
Exponential form ⇄ log form — by = x is exactly the same statement as logbx = y.
Domain of log(x) — only defined for x > 0; any "solution" that breaks this must be rejected.
Extraneous root — an algebraically valid solution that fails the original equation's domain.
Quadratic in disguise — combining log terms sometimes produces a quadratic equation for x.
👁️ Visual aid: switch freely between the log form and exponential form — whichever makes the unknown easier to isolate.
✏️Worked Example 1 — Unknown as the exponentType: Exponent unknown
Question Solve 5(2ˣ) = 80.
Isolate the power
80 ÷ 5 = 16 → 2ˣ = 16 — divide both sides by 5 first, so the unknown power stands alone before logs are taken.Take logs of both sides
log(2ˣ) = log16 — taking the same log of both sides keeps the equation balanced and lets the power law bring x down.Apply the power law and divide
x log2 = log16 → x = log16 ÷ log2 — the power law moves x out in front as a multiplier, then dividing by log2 isolates x.Answer x = 4 (since 2⁴=16 exactly) — a quick check: 2⁴ = 2×2×2×2 = 16 confirms the answer.
🔄Worked Example 2 — Unknown inside a single logType: Inside log, simple
Question Solve log₃(x) = 4.
Convert to exponential form
log₃(x) = 4 → x = 3⁴ — log_b(x) = y is exactly the statement x = bʸ, so rewrite with base 3 raised to the power 4.Evaluate the power
3⁴ = 3×3×3×3 = 81Answer x = 81 — check the domain: x = 81 is positive, so it's a valid solution.
🌡️Worked Example 3 — Combining logs before solvingType: Combine then solve
Question Solve log₂(x) + log₂(x−2) = 3.
Apply the addition law
log₂(x) + log₂(x−2) = log₂[x(x−2)] — two logs of the same base added together combine into the log of a product.Convert to exponential form
log₂[x(x−2)] = 3 → x(x−2) = 2³ = 8 — rewrite the log statement as base 2 raised to the power 3.Open brackets and expand
x(x−2) = x² − 2x, so x² − 2x = 8 → x² − 2x − 8 = 0 — distribute x across the bracket, then move the 8 across to set the equation to zero for factorising.Factorise
(x−4)(x+2) = 0 → x = 4 or x = −2 — find two numbers that multiply to −8 and add to −2.Check the domain x = −2 would make x−2 = −4, and log of a negative number is undefined — this root must be rejected.
Answer x = 4 only (x = −2 rejected: log of a negative number is undefined)
⚙️Worked Example 4 — Engineering context (pH calculation)Type: Engineering
Question pH is defined as pH = −log₁₀[H⁺]. Find [H⁺] for a solution with pH 4.5.
Isolate the log term
4.5 = −log₁₀[H⁺] → log₁₀[H⁺] = −4.5 — multiply both sides by −1 to remove the minus sign in front of the log.Convert from log form to exponential form
log₁₀[H⁺] = −4.5 → [H⁺] = 10−4.5 — log_b(x) = y means x = bʸ, so raise base 10 to the power −4.5.Evaluate
10−4.5 ≈ 3.16×10⁻⁵Answer [H⁺] ≈ 3.16×10⁻⁵ mol/L — the log scale turns a tiny, awkward concentration into the simple number 4.5.
🧠 Common slip: forgetting to check the domain after solving. Any x that makes a log's argument zero or negative must be discarded, however clean the algebra looked.
📈
Graphs of Logarithmic Functions
The shape of y = log_b(x), its asymptote, and why it's the mirror image of the exponential curve.
📖 Key Vocabulary
Asymptote — a line a curve approaches forever but never reaches; here, the y-axis (x = 0).
Domain — the set of x-values allowed in: x > 0 for any log graph.
Range — the set of y-values produced: all real numbers, even though the domain is restricted.
Reflection in y = x — swapping the x and y coordinates of every point on one curve gives the other curve; this is why inverse functions look mirrored.
👁️ Visual aid: the log curve (navy) and the exponential curve (green) are mirror images of each other across the dashed y = x line — swap the coordinates of any point on one and it lands on the other.
✏️Worked Example 1 — Sketching y = log₂(x)Type: Sketching
Question Sketch y = log₂(x), marking any asymptotes and key points.
Build a table of values
x=0.5→y=−1, x=1→y=0, x=2→y=1, x=4→y=2, x=8→y=3 — pick easy powers of 2 so the y-values come out as whole numbers.Mark the asymptote Draw a dashed vertical line at x = 0 — the curve approaches it but never crosses it as x gets small.
Plot and join Plot the points from the table, then draw a smooth increasing curve through them, steepening as it nears the asymptote and flattening as x grows.
Answer A curve rising through (1,0), passing through (2,1), (4,2), (8,3) — climbing fast near x=0, then levelling off.
📐Worked Example 2 — Comparing two basesType: Compare graphs
Question Compare the graphs of y = log₂(x) and y = log₁₀(x). Which rises faster?
Compare at the same x-value
log₂(8) = 3 but log₁₀(8) ≈ 0.903 — the same input gives a much bigger output for the smaller base.Interpret A larger base means each step up the y-axis needs a much bigger jump in x, so the larger-base curve looks flatter.
Answer log₂(x) rises faster (steeper) than log₁₀(x) — both share the point (1,0) and the same asymptote, but the smaller base climbs more quickly.
⚙️Worked Example 3 — Engineering context (reading a dB graph)Type: Engineering
Question A gain-vs-input graph follows L = 20log₁₀(x). Explain why doubling x does not double L.
Apply the shape of the log curve Because the graph flattens as x increases, equal multiplicative jumps in x (like doubling) produce equal-sized steps in L, not proportional ones.
Check with numbers
x=10→L=20, x=20→L≈26, x=40→L≈32 — each doubling of x adds about 6 to L, not double the previous L.Answer The flattening shape of the log graph is exactly why decibel-type scales compress a huge range of x-values into manageable, evenly-spaced steps.
🔧Log Graph Explorer — plot y = log_b(x)Enter a base b and a point x, then press Plot
Enter a base and an x-value, then press Plot to see the point evaluated and marked on the curve…
🧠 Common slip: drawing the log curve as if it starts at the origin (0,0). It never touches the y-axis at all — the curve only crosses the x-axis, at (1,0), and shoots down toward −∞ as x approaches 0.
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🔊 Decibel Scale — Sound & Signal Levels
Levels spanning many orders of magnitude are compressed logarithmically: L = 10log₁₀(P/P₀).
L (dB) = 10 log₁₀(P ÷ P₀)
📟 Try it — Decibel Calculator
Enter P and P₀, then press Calculate…
🧪 pH Scale — Hydrogen Ion Concentration
pH = −log₁₀[H⁺] compresses concentrations spanning 14 orders of magnitude into the familiar 0–14 scale.
pH = −log₁₀[H⁺]
📟 Try it — pH Calculator
Enter the hydrogen ion concentration, then press Calculate…
🌍 Richter Scale — Earthquake Magnitude
Magnitude M = log₁₀(A/A₀) means each whole-number increase represents a tenfold increase in wave amplitude.
M = log₁₀(A ÷ A₀)
📟 Try it — Richter Magnitude
Enter A and A₀, then press Calculate…
📡 Signal Gain — Amplifier dB Gain
Voltage gain in dB is 20log₁₀(V_out/V_in) — the factor of 20 (not 10) arises because power is proportional to voltage squared.
Gain (dB) = 20 log₁₀(V_out ÷ V_in)
📟 Try it — Amplifier Gain
Enter V_in and V_out, then press Calculate…
🎮
Quick-Fire Logarithmic Game
12 questions on the laws and solving logs. How many can you get?
⚡ Logarithmic Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Simplify log 6 + log 4 as a single log.
Addition law: log6 + log4 = log(6×4) = log 24
Question 2 🥉2 marks
Simplify log 100 − log 4 as a single log.
Subtraction law: log100 − log4 = log(100/4) = log 25
Question 3 🥉2 marks
Solve log₂(x) = 5.
Convert to exponential form: log₂(x)=5 → x = 2⁵ = 32
Question 4 🥉2 marks
Express 2log3 as a single logarithm.
Power law: 2log3 = log(3²) = log 9
Question 5 🥈3 marks
Solve 4(3ˣ) = 108 for x.
Isolate the power: 108 ÷ 4 = 27 → 3ˣ = 27
x = 3 (since 3³=27)
x = 3 (since 3³=27)
Question 6 🥈4 marks
Solve log₃(x) + log₃(x+2) = 1.
Addition law + exponential form: log₃[x(x+2)] = 1 → x(x+2) = 3
Open brackets and expand: x²+2x = 3 → x²+2x−3 = 0
Factorise: (x+3)(x−1)=0
x=−3 rejected (log undefined)
x = 1
Open brackets and expand: x²+2x = 3 → x²+2x−3 = 0
Factorise: (x+3)(x−1)=0
x=−3 rejected (log undefined)
x = 1
Question 7 🥈3 marks
Evaluate log₇(50) using change of base, to 3 d.p.
Change of base: log₇50 = ln50 ÷ ln7 = 3.912 ÷ 1.946
≈ 2.010
≈ 2.010
Question 8 🥈4 marks
Solve 2(5ˣ) = 250 for x.
Isolate the power: 250 ÷ 2 = 125 → 5ˣ = 125
x = 3 (since 5³=125)
x = 3 (since 5³=125)
Question 9 🥇4 marks
A jet's sound level is 130 dB. Find the intensity ratio I/I₀.
130 = 10log₁₀(I/I₀)
Divide by 10: log₁₀(I/I₀) = 13
Convert from log: I/I₀ = 10¹³
Divide by 10: log₁₀(I/I₀) = 13
Convert from log: I/I₀ = 10¹³
Question 10 🥇5 marks
Find the pH of a solution with [H⁺] = 2.5×10⁻³ mol/L.
pH = −log₁₀(2.5×10⁻³)
Addition law splits the product: = −(log₁₀2.5 + log₁₀10⁻³)
= −(0.398 − 3)
pH ≈ 2.60
Addition law splits the product: = −(log₁₀2.5 + log₁₀10⁻³)
= −(0.398 − 3)
pH ≈ 2.60
Question 11 🥇5 marks
Two earthquakes have magnitudes 5.2 and 7.2. Find how many times greater the amplitude of the second is.
Difference in magnitude = 7.2 − 5.2 = 2.0
Each whole unit is a ×10 in amplitude, so ratio = 10^(2.0)
= 100 times greater amplitude
Each whole unit is a ×10 in amplitude, so ratio = 10^(2.0)
= 100 times greater amplitude
Question 12 🥇5 marks
An amplifier has V_in = 8 mV and V_out = 3.2 V. Find the voltage gain in dB.
Convert V_in to volts: 8 mV = 0.008 V
Gain = 20log₁₀(3.2/0.008) = 20log₁₀(400)
= 20 × 2.602
≈ 52.0 dB
Gain = 20log₁₀(3.2/0.008) = 20log₁₀(400)
= 20 × 2.602
≈ 52.0 dB