WHAT we do: plug I=I0 into LdB=10log10(I/I0).
WHY: the whole scale is built so the reference point reads zero — this is a sanity anchor.
L=10log10(I0I0)=10log10(1)=10×0=0dB.
The quietest audible sound is the "zero" of the ruler — not silence, but the softest thing your ear can catch.
Recall Solution
pH=−log10(10−7)=−(−7)=7.WHY the minus matters:10−7 has log −7; the minus sign flips it to a friendly positive 7 — the neutral point.
L=10log10(10−1210−5)=10log10(107)=10×7=70dB.WHY 107: dividing powers of ten subtracts exponents, −5−(−12)=7 (Laws of Logarithms, quotient rule). Seven orders of magnitude above threshold → 70 dB.
Recall Solution
M=log10(A0104A0)=log10(104)=4.WHY k=1 here: Richter uses no leading factor, so each whole number is exactly one power of ten of amplitude.
Recall Solution
pH=−log10(2×10−3)=−(log102+log1010−3).WHY split: the product rule turns one awkward log into two easy ones.
=−(0.301+(−3))=−(−2.699)=2.70(strongly acidic).
WHAT: convert each level to intensity, find the ratio of total to one.
WHY: decibels never add directly — physical intensities do (see the +3 dB rule).
One whisper: I1=I01030/10=I0103. The crowd: Ic=I01050/10=I0105.
n=I1Ic=103105=102=100whispers.
A jump of +20 dB is a ×100 jump in intensity: ΔL=10log10r⇒20=10log10r⇒r=102.
Recall Solution
[H+]B[H+]A=10−610−3=10−3−(−6)=103=1000times.WHY subtract exponents: the quotient rule of logs in reverse. A 3-unit pH gap is a factor of 103 in acidity — the minus signs cancel neatly.
Recall Solution
A5.0A6.5=106.5−5.0=101.5.WHY 101.5: subtract magnitudes, then exponentiate. Now evaluate 101.5=10⋅100.5=1010≈10×3.162=31.6times larger.
WHAT: four identical sources → intensity ×4.
WHY: intensities add, and four equal ones give 4I.
ΔL=10log10(4)=10×0.602=6.02dB.
New level =80+6.02=86.0dB.
Cross-check with the doubling rule: ×4 is two doublings, +3 dB then +3 dB =+6 dB. ✓ The picture below shows how each doubling adds a fixed chalk-step.
Recall Solution
WHAT: average the concentrations (linear quantities), then take the log.
WHY: you cannot average pH values — pH is a logarithm, and log of an average ≠ average of logs.
[H+]mix=210−2+10−4=20.01+0.0001=20.0101=5.05×10−3.pH=−log10(5.05×10−3)=−(log105.05−3)=−(0.703−3)=2.30.
Notice the mixture is close to pH 2, not the naive average pH 3 — the stronger acid dominates because it has 100× more ions.
Recall Solution
ΔL=78−90=−12dB=10log10(r)⇒log10(r)=−1.2⇒r=10−1.2.WHY negative: the level dropped, so r<1.
r=10−1.2=10−2⋅100.8≈0.01×6.31=0.0631≈6.3%.
So about 93.7% of the intensity was absorbed — a 12 dB drop keeps only ~1/16 of the sound.
WHAT: distance ×10 → intensity ×1021=1001.
WHY inverse-square: the same power spreads over a sphere of area ∝d2, so intensity ∝1/d2.
ΔL=10log10(1001)=10×(−2)=−20dB.
New level =100−20=80dB.
Key insight: every 10× in distance costs a fixed −20 dB, no matter the starting level — the step rule ΔL=klog10r makes distance behave logarithmically too. The figure traces this fall.
Recall Solution
WHAT: subtract the log-energies.
WHY:log10E=1.5M+c, so the constant c cancels in a difference.
log10E8−log10E6=1.5(8)−1.5(6)=1.5(8−6)=1.5×2=3.E6E8=103=1000timesmore energy.
Note: the amplitude ratio is only 108−6=100, but energy grows as the 1.5-power of the magnitude gap, giving 1000. This is why a "2-point" quake feels catastrophically worse.
Recall Solution
(a) SNR in dB =65−45=20 dB. As an intensity ratio: r=1020/10=102=100:1.
WHY: the dB difference is directly 10log10(ratio), so invert it.
(b) Two identical noise sources → noise intensity ×2 → noise level rises by 10log102≈3.01 dB.
New noise =45+3.01=48.0 dB. Signal unchanged at 65 dB.
New SNR =65−48.01=16.99dB≈17dB, i.e. it dropped by about 3 dB — exactly the doubling penalty. The extra noise source halved your effective clarity.
Recall Full-page self-test (no notes)
Level 0 dB — is that silence? (No: I=I0.)
Two 60 dB sources combine to? (63 dB, +3 for doubling.)
pH 2 vs pH 5 acidity ratio? (103=1000.)
Magnitude 8 vs 6 energy factor? (103=1000.)
Distance ×10 changes dB by? (−20, via inverse-square.)
