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IB Mathematics: Applications and Interpretation HL · 12.6: Exponents & Logarithms · 12.6.4

Using logarithmic scales on graphs Problem set A Answers

Skills

    1. (a) 0.602, 2.40, 4.26, 4.91
    2. (b) 1
    1. (a) gradient 0.300, intercept 0.900
    2. (b) a = 7.94, b = 2.00
    3. (c) y = 3980
    1. (a) gradient 0.900, intercept 1.30
    2. (b) k = 0.900, a = 20.0
    3. (c) y = 675
    1. (a) 0.778, 1.18, 1.57, 1.97
    2. (b) Each step adds 0.398: log y against x is a straight line
    3. (c) y = 6 × 2.5^x

In context

    1. (a) gradient 0.163, intercept 0.378, r = 0.9994
    2. (b) N = 2.39 × 1.45^t
    3. (c) N = 101
    1. (a) gradient 2.98, intercept −0.243, r = 0.9999
    2. (b) P = 0.572 × v^2.98
    3. (c) P = 1490 kW

Stretch

    1. (a) r = 0.9056
    2. (b) r = 0.9999
    3. (c) The power model; y = 2.41 × x^1.60
    1. (a) log10 y = 1.08 − 0.0969x
    2. (b) gradient −0.0969, intercept 1.08
    3. (c) log10 y = 0.699 + 3 log10 x; gradient 3, intercept 0.699

jmaths.xyz · AI HL · 12.6.4 Logarithmic scales on graphs