Find the value of a and of b.

Use the regression equation to estimate the value of y when x = 3.57.

The relationship between x and y can be modelled using the formula y = kxⁿ, where k ≠ 0 , n ≠ 0 , n ≠ 1.

By expressing ln y in terms of ln x, find the value of n and of k.

valid approach      (M1)

eg  one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14      A1A1 N3

[3 marks]

correct substitution     (A1)

eg   −0.454 ln 3.57 + 6.14

correct working     (A1)

eg  ln y = 5.56484

261.083 (260.409 from 3 sf)

y = 261, (y = 260 from 3sf)       A1 N3

Note: If no working shown, award N1 for 5.56484.
If no working shown, award N2 for ln y = 5.56484.

[3 marks]

METHOD 1

valid approach for expressing ln y in terms of ln x      (M1)

eg  \({\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b\)

correct application of addition rule for logs      (A1)

eg  \({\text{ln}}\,k + {\text{ln}}\,\left( {{x^n}} \right)\)

correct application of exponent rule for logs       A1

eg  \({\text{ln}}\,k + n\,{\text{ln}}\,x\)

comparing one term with regression equation (check FT)      (M1)

eg  \(n = a,\,\,b = {\text{ln}}\,k\)

correct working for k      (A1)

eg  \({\text{ln}}\,k = 6.14210,\,\,\,k = {e^{6.14210}}\)

465.030

\(n =  - 0.454,\,\,k = 465\) (464 from 3sf)     A1A1 N2N2

 

METHOD 2

valid approach      (M1)

eg  \({e^{{\text{ln}}\,y}} = {e^{a\,{\text{ln}}\,x + b}}\)

correct use of exponent laws for \({e^{a\,{\text{ln}}\,x + b}}\)     (A1)

eg  \({e^{a\,{\text{ln}}\,x}} \times {e^b}\)

correct application of exponent rule for \(a\,{\text{ln}}\,x\)     (A1)

eg  \({\text{ln}}\,{x^a}\)

correct equation in y      A1

eg  \(y = {x^a} \times {e^b}\)

comparing one term with equation of model (check FT)      (M1)

eg  \(k = {e^b},\,\,n = a\)

465.030

\(n =  - 0.454,\,\,k = 465\) (464 from 3sf)     A1A1 N2N2

 

METHOD 3

valid approach for expressing ln y in terms of ln x (seen anywhere)      (M1)

eg  \({\text{ln}}\,y = {\text{ln}}\,\left( {k{x^n}} \right),\,\,{\text{ln}}\,\left( {k{x^n}} \right) = a\,{\text{ln}}\,x + b\)

correct application of exponent rule for logs (seen anywhere)      (A1)

eg  \({\text{ln}}\,\left( {{x^a}} \right) + b\)

correct working for b (seen anywhere)      (A1)

eg  \(b = {\text{ln}}\,\left( {{e^b}} \right)\)

correct application of addition rule for logs      A1

eg  \({\text{ln}}\,\left( {{e^b}{x^a}} \right)\)

comparing one term with equation of model (check FT)     (M1)

eg  \(k = {e^b},\,\,n = a\)

465.030

\(n =  - 0.454,\,\,k = 465\) (464 from 3sf)     A1A1 N2N2

[7 marks]

[N/A]

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