Home / IB DP Math AI: Topic : SL 1.5: Laws of exponents: IB style Questions SL Paper 2

IB DP Math AI: Topic : SL 1.5: Laws of exponents: IB style Questions SL Paper 2

Question

The following table shows values of ln x and ln y.

The relationship between ln x and ln y can be modelled by the regression equation ln y = a ln x + b.

Find the value of a and of b.[3]

a.

Use the regression equation to estimate the value of y when x = 3.57.[3]

b.

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.[7]

c.
Answer/Explanation

Markscheme

valid approach (M1)

eg one correct value

−0.453620, 6.14210

a = −0.454, b = 6.14 A1A1 N3

[3 marks]

a.

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]

b.

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]

c.

Question

Expand \({(x – 2)^4}\) and simplify your result.

[3]
a.

Find the term in \({x^3}\) in \((3x + 4){(x – 2)^4}\) .

[3]
b.
Answer/Explanation

Markscheme

evidence of expanding     M1

e.g. \({(x – 2)^4} = {x^4} + 4{x^3}( – 2) + 6{x^2}{( – 2)^2} + 4x{( – 2)^3} + {( – 2)^4}\)     A2     N2

\({(x – 2)^4} = {x^4} – 8{x^3} + 24{x^2} – 32x + 16\)

[3 marks]

a.

finding coefficients, \(3 \times 24( = 72)\) , \(4 \times( – 8)( = – 32)\)     (A1)(A1)

term is \(40{x^3}\)     A1     N3

[3 marks]

b.

Question

Let \(f(x) = {\log _3}\frac{x}{2} + {\log _3}16 – {\log _3}4\) , for \(x > 0\) .

Show that \(f(x) = {\log _3}2x\) .

[2]
a.

Find the value of \(f(0.5)\) and of \(f(4.5)\) .

[3]
b.

The function f can also be written in the form \(f(x) = \frac{{\ln ax}}{{\ln b}}\) .

(i)     Write down the value of a and of b .

(ii)    Hence on graph paper, sketch the graph of f , for \( – 5 \le x \le 5\) , \( – 5 \le y \le 5\) , using a scale of 1 cm to 1 unit on each axis.

(iii)   Write down the equation of the asymptote.

[6]
c(i), (ii) and (iii).

Write down the value of \({f^{ – 1}}(0)\) .

[1]
d.

The point A lies on the graph of f . At A, \(x = 4.5\) .

On your diagram, sketch the graph of \({f^{ – 1}}\) , noting clearly the image of point A.

[4]
e.
Answer/Explanation

Markscheme

combining 2 terms     (A1)

e.g. \({\log _3}8x – {\log _3}4\) , \({\log _3}\frac{1}{2}x + {\log _3}4\)

expression which clearly leads to answer given     A1

e.g. \({\log _3}\frac{{8x}}{4}\) , \({\log _3}\frac{{4x}}{2}\)

\(f(x) = {\log _3}2x\)     AG     N0

[2 marks]

a.

attempt to substitute either value into f     (M1)

e.g. \({\log _3}1\) , \({\log _3}9\)

\(f(0.5) = 0\) , \(f(4.5) = 2\)     A1A1     N3

[3 marks]

b.

(i) \(a = 2\) , \(b = 3\)     A1A1     N1N1

(ii)
     A1A1A1     N3

Note: Award A1 for sketch approximately through \((0.5 \pm 0.1{\text{, }}0 \pm 0.1)\) , A1 for approximately correct shape, A1 for sketch asymptotic to the y-axis.

(iii) \(x = 0\) (must be an equation)     A1     N1

[6 marks]

c(i), (ii) and (iii).

\({f^{ – 1}}(0) = 0.5\)     A1     N1

[1 mark]

d.


     A1A1A1A1     N4

Note: Award A1 for sketch approximately through \((0 \pm 0.1{\text{, }}0.5 \pm 0.1)\) , A1 for approximately correct shape of the graph reflected over \(y = x\) , A1 for sketch asymptotic to x-axis, A1 for point \((2 \pm 0.1{\text{, }}4.5 \pm 0.1)\) clearly marked and on curve.

[4 marks]

e.
Scroll to Top