Year 10 Mathematics | Victorian Curriculum 2.0
Logarithmic scales
Topic 10 | Measurement | Answer key

Tier 1

    1. log⁡101000=3\log_{10} 1000 = 3log10​1000=3, since 103=100010^3 = 1000103=1000.
    2. log⁡100.01=−2\log_{10} 0.01 = -2log10​0.01=−2, since 10−2=0.0110^{-2} = 0.0110−2=0.01.
    3. x=104=10 000x = 10^4 = 10\,000x=104=10000.
    4. 5 000 000500=10 000=104\dfrac{5\,000\,000}{500} = 10\,000 = 10^45005000000​=10000=104. They are 444 orders of magnitude apart.
    5. 1030/10=103=100010^{30/10} = 10^3 = 10001030/10=103=1000 times more intense.
    6. Amplitude ratio: 106.0−4.0=102=10010^{6.0 - 4.0} = 10^2 = 100106.0−4.0=102=100 times larger.
    7. pH=−log⁡10(10−3)=3\text{pH} = -\log_{10}(10^{-3}) = 3pH=−log10​(10−3)=3.
    8. 107−5=102=10010^{7-5} = 10^2 = 100107−5=102=100 times greater.

Tier 2

    1. (a) Amplitude ratio: 106.5−4.5=102=10010^{6.5 - 4.5} = 10^2 = 100106.5−4.5=102=100 times. (b) Energy ratio: 31.62≈100031.6^2 \approx 100031.62≈1000 times.
    2. Difference: 75−20=5575 - 20 = 5575−20=55 dB. Intensity ratio: 1055/10=105.5≈316 22810^{55/10} = 10^{5.5} \approx 316\,2281055/10=105.5≈316228 times more intense.
    3. The values span 444 orders of magnitude (222 to 20 00020\,00020000). On a linear scale, 222 and 202020 would be indistinguishable near the axis while 20 00020\,00020000 dominates. A log scale spaces all five points evenly, revealing the constant factor-of-101010 pattern.
    4. Coffee is more acidic (lower pH). Concentration ratio: 1011.5−5=106.5≈3 162 27810^{11.5 - 5} = 10^{6.5} \approx 3\,162\,2781011.5−5=106.5≈3162278 times more hydrogen ions in the coffee.
    5. After 505050 years (5 doublings): 5000×25=160 0005000 \times 2^5 = 160\,0005000×25=160000. On a log scale, exponential growth (constant doubling time) appears as a straight line because log⁡(P)=log⁡(5000)+t×log⁡210\log(P) = \log(5000) + t \times \frac{\log 2}{10}log(P)=log(5000)+t×10log2​, which is linear in ttt.
    6. Ratio =103=1000= 10^3 = 1000=103=1000.

Tier 3

    1. log⁡100\log_{10} 0log10​0 is undefined because there is no power nnn such that 10n=010^n = 010n=0 (powers of 101010 are always positive). On a log-scale graph, zero cannot be plotted — the axis extends toward −∞-\infty−∞ in log-space as values approach zero. This means log scales can only represent strictly positive data.
    2. Each magnitude step is a factor of 2.5122.5122.512. Over 555 steps: 2.5125≈1002.512^5 \approx 1002.5125≈100. A magnitude-111 star is about 100100100 times brighter than a magnitude-666 star.
    3. The Richter scale is logarithmic, not linear. A magnitude 888 quake has 108−4=104=10 00010^{8-4} = 10^4 = 10\,000108−4=104=10000 times the wave amplitude of a magnitude 444 quake — not 222 times. The student confused additive and multiplicative differences.
    4. Example table (masses): electron ≈10−30\approx 10^{-30}≈10−30 kg, grain of sand ≈10−6\approx 10^{-6}≈10−6 kg, human ≈102\approx 10^2≈102 kg, Earth ≈1024\approx 10^{24}≈1024 kg, Sun ≈1030\approx 10^{30}≈1030 kg. These span about 606060 orders of magnitude. A log scale is essential because a linear axis from 10−3010^{-30}10−30 to 103010^{30}1030 would make all but the largest value invisible.

Challenge

    1. For M=5.0M = 5.0M=5.0: log⁡10E=1.5(5)+4.8=12.3\log_{10} E = 1.5(5) + 4.8 = 12.3log10​E=1.5(5)+4.8=12.3, so E≈2.0×1012E \approx 2.0 \times 10^{12}E≈2.0×1012 J. For M=8.0M = 8.0M=8.0: log⁡10E=1.5(8)+4.8=16.8\log_{10} E = 1.5(8) + 4.8 = 16.8log10​E=1.5(8)+4.8=16.8, so E≈6.3×1016E \approx 6.3 \times 10^{16}E≈6.3×1016 J. Ratio: 6.3×10162.0×1012≈31 500≈31.63\dfrac{6.3 \times 10^{16}}{2.0 \times 10^{12}} \approx 31\,500 \approx 31.6^32.0×10126.3×1016​≈31500≈31.63 (since 31.63≈31 62331.6^3 \approx 31\,62331.63≈31623). Confirmed.
    2. Each source has intensity I=I0×1080/10=108I0I = I_0 \times 10^{80/10} = 10^8 I_0I=I0​×1080/10=108I0​. Combined intensity =2×108I0= 2 \times 10^8 I_0=2×108I0​. Combined level =10log⁡10(2×108)=10(log⁡102+8)≈10(0.301+8)=83.01= 10\log_{10}(2 \times 10^8) = 10(\log_{10} 2 + 8) \approx 10(0.301 + 8) = 83.01=10log10​(2×108)=10(log10​2+8)≈10(0.301+8)=83.01 dB. Doubling intensity adds about 333 dB, not 808080 dB.
    3. (a) log⁡10 ⁣(108102)=log⁡10(106)=6\log_{10}\!\left(\frac{10^8}{10^2}\right) = \log_{10}(10^6) = 6log10​(102108​)=log10​(106)=6 orders of magnitude. (b) A straight line, because log⁡(count)\log(\text{count})log(count) increases linearly with time for exponential growth. (c) Total growth factor =106= 10^6=106 over 242424 hours. Hourly factor =(106)1/24=100.25≈1.778= (10^6)^{1/24} = 10^{0.25} \approx 1.778=(106)1/24=100.25≈1.778.
    4. If M0M_0M0​ increases by a factor of 1000=1031000 = 10^31000=103, then log⁡10(M0)\log_{10}(M_0)log10​(M0​) increases by 333. So MwM_wMw​ increases by 23×3=2\frac{2}{3} \times 3 = 232​×3=2 units.
Year 10 Mathematics study companion | Answer key