Logarithmic scale — decibels, Richter, pH: Worked examples

Everything below uses only three master facts from the parent note. Say them once, out loud:

The scenario matrix

Every log-scale problem falls into one of these cells. The last column names the example that defends it.

#	Case class	What makes it tricky	Covered by
A	Forward, plain — given quantity, find scale value	choosing the right $k$	Ex 1
B	Backward, plain — given scale value, find quantity	inverting the log	Ex 2
C	Ratio/step — "how many times bigger?"	subtract then exponentiate	Ex 3
D	Adding sources — combining two loudnesses	add intensities, never dB	Ex 4
E	Negative-log scale (pH) — the minus sign	sign flips the direction	Ex 5
F	Fractional scale value — non-integer pH / dB	$1 0^{0.5} = 10$ etc.	Ex 6
G	Degenerate: quantity = reference ( $I = I_{0}$ )	$lo g_{10} 1 = 0$	Ex 7
H	Limiting: quantity → 0	$lo g_{10} (tiny) \to - \infty$	Ex 7
I	Real-world word problem — reverse-engineer a factor	translate words to $r$	Ex 8
J	Exam twist — mix two scales / hidden step	don't panic at the wrapping	Ex 9

We will hit all ten cells in nine examples (Ex 7 covers both G and H — the two edge cases live together on one picture).

Figure — Logarithmic scale — decibels, Richter, pH

The figure above is your map: read the horizontal axis as the raw quantity (multiplying by 10 each big tick) and the vertical axis as the scale value (rising by a constant step). Every example below is a point, an arrow, or a gap on this one picture.

Cell A — Forward, plain

Worked example Ex 1 · Sound intensity → decibels

A vacuum cleaner produces intensity $I = 1 0^{- 4} W/m^{2}$ . The threshold of hearing is $I_{0} = 1 0^{- 12} W/m^{2}$ . Find the sound level in dB.

Forecast: Guess first — is the answer closer to 8, 80, or 800 dB? Jot it down.

Write the decibel formula. Why this step? Cell A means we have the quantity and want the scale value, so we use the forward fact with $k = 10$ . $L = 10 lo g_{10} (\frac{I}{I _{0}}) .$
Form the ratio $\frac{I}{I _{0}} = \frac{1 0 ^{- 4}}{1 0 ^{- 12}} = 1 0^{- 4 - (- 12)} = 1 0^{8}$ . Why? The quotient rule of logs is easiest when the ratio is a clean power of 10 — subtract exponents (see Laws of Logarithms).
$L = 10 lo g_{10} (1 0^{8}) = 10 \times 8 = 80 dB$ . Why? $lo g_{10} (1 0^{8})$ asks "10 to what power is $1 0^{8}$ ?" — answer $8$ .

Verify: Units: $I / I_{0}$ is dimensionless (W/m² over W/m²), so $lo g$ is legal and dB is unitless. A vacuum cleaner is around 80 dB in real life. ✓ (Did your forecast say 80?)

Cell B — Backward, plain

Cell C — Ratio / step

Cell D — Adding sources

Worked example Ex 4 · Three identical drummers

One drummer measures $70 dB$ . Three identical drummers play together. New level?

Forecast: More than 210 dB? (Trap!) Or barely above 70?

Convert to intensity. Each drummer: $I = I_{0} 1 0^{70/10} = I_{0} 1 0^{7}$ . Why this step? You cannot add decibels (Cell D's whole trap). Physical intensities add; decibels are logarithms of them.
Total intensity $I_{tot} = 3 \times I_{0} 1 0^{7}$ . Why? Three equal, independent sources deliver three times the power to your ear.
Back to dB: $L = 10 lo g_{10} (\frac{3 \times 1 0 ^{7} I _{0}}{I _{0}}) = 10 lo g_{10} (3 \times 1 0^{7})$ . Why? Forward fact again, now with the combined intensity.
Split with the product rule: $L = 10 (lo g_{10} 3 + 7) = 10 (0.4771 + 7) = 74.77 \approx 74.8 dB$ . Why? $lo g (3 \times 1 0^{7}) = lo g 3 + lo g 1 0^{7}$ (Laws of Logarithms).

Verify: Tripling intensity should add $10 lo g_{10} 3 \approx 4.77 dB$ on top of 70 → $74.8$ . ✓ Notice: three times the sound is under 5 dB louder — that's the log scale crushing the range.

Cell E — Negative-log scale (pH)

Cell F — Fractional scale value

Cells G & H — Degenerate and limiting inputs

Both edge cases live on one picture. Look at where the curve crosses zero and where it dives off the bottom.

Worked example Ex 7 · When quantity equals the reference, and when it vanishes

(G) Degenerate: A sound has intensity exactly equal to the threshold, $I = I_{0}$ . What is $L$ ?

$L = 10 lo g_{10} (I_{0} / I_{0}) = 10 lo g_{10} (1)$ . Why this step? Cell G tests the reference point — the value the whole scale is measured from.
$lo g_{10} (1) = 0$ because $1 0^{0} = 1$ . So $L = 10 \times 0 = 0 dB$ . Why? The reference is defined to sit at zero on the scale — that's what "reference" means.

(H) Limiting: What happens to the pH as $[H^{+}] \to 0$ (pure, ion-free water in the ideal limit)?

$pH = - lo g_{10} [H^{+}]$ . As $[H^{+}]$ shrinks toward $0$ , its log dives toward $- \infty$ , so $- lo g_{10} [H^{+}] \to + \infty$ . Why this matters? The log scale has no top and no bottom of quantity except zero — you can never reach quantity $= 0$ on the scale; you only run off to infinity. Look at the red dashed asymptote in the figure: the curve hugs the vertical axis but never touches it.

Verify: Numerically, $[H^{+}] = 1 0^{- 14}$ gives pH $14$ ; $1 0^{- 20}$ gives pH $20$ — it keeps climbing without limit. ✓ (And $lo g_{10} (1) = 0$ exactly.) This is why $0 dB$ is "the quietest audible sound," not "silence."

Cell I — Real-world word problem

Worked example Ex 8 · Noise regulation

A factory must cut its noise power to one-quarter of its current value to meet code. By how many dB does the reading drop?

Forecast: More or less than a 4 dB drop?

Translate the words: "one-quarter the power" means the multiplier $r = \frac{1}{4} = 0.25$ . Why this step? Cell I is about turning English into a ratio $r$ , then the step rule does the rest.
Apply $Δ L = k lo g_{10} r$ with $k = 10$ : $Δ L = 10 lo g_{10} (0.25)$ . Why? The change depends only on the factor, not on the starting level (the parent's key result).
$lo g_{10} (0.25) = lo g_{10} (1/4) = - lo g_{10} 4 = - 0.602$ . So $Δ L = 10 \times (- 0.602) = - 6.0 dB$ . Why? Quotient rule turns $1/4$ into a negative log; a drop correctly comes out negative.

Verify: Cross-check with the "+3 dB per ×2" fact: quarter = two halvings = $- 3 - 3 = - 6 dB$ . ✓ The two methods agree.

Cell J — Exam-style twist (two scales at once)

Worked example Ex 9 · The hidden step

A quiet library is at pH-style thinking… no — here's the real twist. An earthquake of magnitude $M_{1}$ has 1000 times the ground amplitude of one at magnitude $M_{2}$ . Then a separate sound doubles in intensity. Combine both facts into scale changes.

Forecast: Will the Richter difference be 3, and the dB change 3 as well — coincidence?

Richter part. Amplitude ratio $r = 1000 = 1 0^{3}$ , with $k = 1$ : $Δ M = lo g_{10} (1 0^{3}) = 3$ . So the magnitudes differ by exactly $3$ . Why this step? Cell J hides two independent applications of the same step rule inside one sentence; peel them apart.
Decibel part. Intensity ratio $r = 2$ , with $k = 10$ : $Δ L = 10 lo g_{10} (2) = 10 \times 0.301 = 3.01 dB$ . Why the different constant? Same rule $Δ = k lo g_{10} r$ , but $k = 10$ for sound versus $k = 1$ for Richter — the constant $k$ is the only thing that changes between scales.
Note the "coincidence": $lo g_{10} (1 0^{3}) = 3$ and $10 lo g_{10} 2 \approx 3$ both give $\approx 3$ , but for completely different reasons — one is an exact power of ten, the other is a doubling scaled by ten. Why point this out? Exams love numbers that look equal to trap you into merging unrelated scales.

Verify: Richter: $lo g_{10} (1000) = 3$ exactly. dB: $10 lo g_{10} 2 = 3.0103$ . Both confirmed independently. ✓

Recall One-line recipe for any cell

Ask two questions: (1) Do I have the quantity or the scale value? → pick forward $L = k lo g_{10} (I / I_{0})$ or backward $I = I_{0} 1 0^{L / k}$ . (2) Am I asked for a ratio/change? → use $Δ L = k lo g_{10} r$ . Then just plug in $k$ ( $10$ dB, $1$ Richter, $- 1$ pH). Every example above is one of these two moves.

Active recall

Redo without the answers, then check:

Ex 1 answer (vacuum,

1 0^{- 4}

W/m²)

80 dB

, since

lo g_{10} (1 0^{8}) = 8

Ex 3 answer (85 dB vs 61 dB intensity ratio)

1 0^{2.4} \approx 251

times.

Ex 4 answer (three 70-dB drummers)

\approx 74.8 dB

(add intensities:

10 lo g_{10} 3 \approx 4.77

up).

Ex 5 answer (pH of

2 \times 1 0^{- 3}

)

pH \approx 2.70

Ex 6 answer (concentration at pH 4.5)

3.16 \times 1 0^{- 5} mol/L

(that

10

factor).

Ex 7 answer (

I = I_{0}

, and

[H^{+}] \to 0

)

0 dB

; pH

\to + \infty

Ex 8 answer (quarter the power)

drop of

6.0 dB

(two halvings).

Ex 9 answers (1000× amplitude; doubled intensity)

Δ M = 3

exactly;

Δ L \approx 3.01 dB

Connections

Laws of Logarithms — every "split the log" step is the product/quotient rule.
Exponential Functions — the $1 0^{(\cdot)}$ that un-logs a scale value (Cells B, F).
Change of Base Formula — why base-10 keeps these ratios readable.
Orders of Magnitude — "×10 is one step" is the beating heart of every cell.
Weber–Fechner Law — why perception itself is a log scale (Cell D's intuition).