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

Using logarithmic scales on graphs Problem set B Answers

Skills

    1. (a) −1.22, 1.95, 3.88, 6.68
    2. (b) −2
    1. (a) gradient 0.100, intercept 1.40
    2. (b) a = 25.1, b = 1.26
    3. (c) y = 794
    1. (a) gradient 0.600, intercept 0.500
    2. (b) k = 0.600, a = 3.16
    3. (c) y = 76.0
    1. (a) 1.08, 1.28, 1.49, 1.69
    2. (b) Each step adds 0.204: log y against x is a straight line
    3. (c) y = 12 × 1.6^x

In context

    1. (a) gradient −0.0861, intercept 2.98, r = −0.9986
    2. (b) N = 952 × 0.820^t
    3. (c) N = 48.7
    1. (a) gradient 0.498, intercept 0.302, r = 0.9990
    2. (b) T = 2.00 × L^0.498
    3. (c) T = 6.31 s

Stretch

    1. (a) r = 0.9998
    2. (b) r = 0.9664
    3. (c) The exponential model; y = 5.04 × 1.90^x
    1. (a) log10 y = 1.60 − 0.187x
    2. (b) gradient −0.187, intercept 1.60
    3. (c) log10 y = 0.903 + 2 log10 x; gradient 2, intercept 0.903

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