# IB Math Analysis & Approaches Questionbank-Topic: SL 2.9 Exponential and Logarithmic functions-SL Paper 1

### Question

Dominic jumps out of an airplane that is flying at constant altitude. Before opening his

parachute, he goes through a period of freefall.

Dominic’s vertical speed during the time of freefall, S , in m s1 , is modelled by the following function.

$$S(t)=K-6-(1.2^t),t\geqslant 0$$

where t , is the number of seconds after he jumps out of the airplane, and K is a constant. A sketch

of Dominic’s vertical speed against time is shown below. Dominic’s initial vertical speed is $$0\:ms^{-1}$$

1. Find the value of K . 

2. In the context of the model, state what the horizontal asymptote represents. 

3. Find Dominic’s vertical speed after 10 seconds. Give your answer in km h−1. 

Ans:

(a)

0 = K – 60(1.20)

K= 60

(b)

the (vertical) speed that Dominic is approaching (as t increases)

OR

the limit of the (vertical) speed of Dominic

(c)

S=60-60$$(1.2^{-10})$$

$$S=50.3096..(ms^{-1})$$

$$181(kmh^{-1})(181.144..(kmh^{-1}))$$

### Question

Let f (x)a log3 (x – 4) , for x > 4 , where a > 0 .

Point A(13 , 7) lies on the graph of f .

1. Find the value of a . 

The x-intercept of the graph of f is (5 , 0) .

2. On the following grid, sketch the graph of f . Ans:

(a)

attempt to substitute coordinates (in any order) into f

eg a $$log_{3} (13-4)=7, alog_{3}$$(7-4)=13 , alog9 = 7

finding $$log_{3}$$ 9=2 (seen anywhere)

eg $$log_{3}9=2, 2a=7$$

$$a= \frac{7}{2}$$

(b) ### Question

Let $$f(x) = {\log _p}(x + 3)$$ for $$x > – 3$$ . Part of the graph of f is shown below. The graph passes through A(6, 2) , has an x-intercept at (−2, 0) and has an asymptote at $$x = – 3$$ .

Find p .


a.

The graph of f is reflected in the line $$y = x$$ to give the graph of g .

(i)     Write down the y-intercept of the graph of g .

(ii)    Sketch the graph of g , noting clearly any asymptotes and the image of A.


b.

The graph of $$f$$ is reflected in the line $$y = x$$ to give the graph of $$g$$ .

Find $$g(x)$$ .


c.

## Markscheme

evidence of substituting the point A     (M1)

e.g. $$2 = {\log _p}(6 + 3)$$

manipulating logs     A1

e.g. $${p^2} = 9$$

$$p = 3$$     A2     N2

[4 marks]

a.

(i) $$y = – 2$$ (accept $$(0{\text{, }} – 2))$$     A1     N1

(ii) A1A1A1A1     N4

Note: Award A1 for asymptote at $$y = – 3$$ , A1 for an increasing function that is concave up, A1 for a positive x-intercept and a negative y-intercept, A1 for passing through the point $$(2{\text{, }}6)$$ .

[5 marks]

b.

METHOD 1

recognizing that $$g = {f^{ – 1}}$$     (R1)

evidence of valid approach     (M1)

e.g. switching x and y (seen anywhere), solving for x

correct manipulation     (A1)

e.g. $${3^x} = y + 3$$

$$g(x) = {3^x} – 3$$     A1     N3

METHOD 2

recognizing that $$g(x) = {a^x} + b$$     (R1)

identifying vertical translation     (A1)

e.g. graph shifted down 3 units, $$f(x) – 3$$

evidence of valid approach     (M1)

e.g. substituting point to identify the base

$$g(x) = {3^x} – 3$$     A1     N3

[4 marks]

c.

## Question

Find the value of $${\log _2}40 – {\log _2}5$$ .


a.

Find the value of $${8^{{{\log }_2}5}}$$ .


b.

## Markscheme

evidence of correct formula    (M1)

eg   $$\log a – \log b = \log \frac{a}{b}$$ , $$\log \left( {\frac{{40}}{5}} \right)$$ , $$\log 8 + \log 5 – \log 5$$

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   $${\log _2}8$$ , $${2^3} = 8$$

$${\log _2}40 – {\log _2}5 = 3$$     A1     N2

[3 marks]

a.

attempt to write $$8$$ as a power of $$2$$ (seen anywhere)     (M1)

eg   $${({2^3})^{{{\log }_2}5}}$$ , $${2^3} = 8$$ , $${2^a}$$

multiplying powers     (M1)

eg   $${2^{3{{\log }_2}5}}$$ , $$a{\log _2}5$$

correct working     (A1)

eg   $${2^{{{\log }_2}125}}$$ , $${\log _2}{5^3}$$ , $${\left( {{2^{{{\log }_2}5}}} \right)^3}$$

$${8^{{{\log }_2}5}} = 125$$     A1     N3

[4 marks]

b.

## Question

Let $$f(x) = {{\rm{e}}^{x + 3}}$$ .

(i)     Show that $${f^{ – 1}}(x) = \ln x – 3$$ .

(ii)    Write down the domain of $${f^{ – 1}}$$ .


a.

Solve the equation $${f^{ – 1}}(x) = \ln \frac{1}{x}$$ .


b.

## Markscheme

(i) interchanging x and y (seen anywhere)     M1

e.g. $$x = {{\rm{e}}^{y + 3}}$$

correct manipulation     A1

e.g. $$\ln x = y + 3$$ , $$\ln y = x + 3$$

$${f^{ – 1}}(x) = \ln x – 3$$     AG     N0

(ii) $$x > 0$$     A1     N1

[3 marks]

a.

collecting like terms; using laws of logs     (A1)(A1)

e.g. $$\ln x – \ln \left( {\frac{1}{x}} \right) = 3$$ , $$\ln x + \ln x = 3$$ , $$\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3$$ , $$\ln {x^2} = 3$$

simplify     (A1)

e.g. $$\ln x = \frac{3}{2}$$ ,  $${x^2} = {{\rm{e}}^3}$$

$$x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)$$     A1     N2

[4 marks]

b.

## Examiners report

Many candidates interchanged the $$x$$ and $$y$$ to find the inverse function, but very few could write down the correct domain of the inverse, often giving $$x \ge 0$$ , $$x > 3$$ and “all real numbers” as responses.

a.

Where students attempted to solve the equation in (b), most treated $$\ln x – 3$$ as $$\ln (x – 3)$$ and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.

b.

## Question

Find the value of $${\log _2}40 – {\log _2}5$$ .


a.

Find the value of $${8^{{{\log }_2}5}}$$ .


b.

## Markscheme

evidence of correct formula    (M1)

eg   $$\log a – \log b = \log \frac{a}{b}$$ , $$\log \left( {\frac{{40}}{5}} \right)$$ , $$\log 8 + \log 5 – \log 5$$

Note: Ignore missing or incorrect base.

correct working     (A1)

eg   $${\log _2}8$$ , $${2^3} = 8$$

$${\log _2}40 – {\log _2}5 = 3$$     A1     N2

[3 marks]

a.

attempt to write $$8$$ as a power of $$2$$ (seen anywhere)     (M1)

eg   $${({2^3})^{{{\log }_2}5}}$$ , $${2^3} = 8$$ , $${2^a}$$

multiplying powers     (M1)

eg   $${2^{3{{\log }_2}5}}$$ , $$a{\log _2}5$$

correct working     (A1)

eg   $${2^{{{\log }_2}125}}$$ , $${\log _2}{5^3}$$ , $${\left( {{2^{{{\log }_2}5}}} \right)^3}$$

$${8^{{{\log }_2}5}} = 125$$     A1     N3

[4 marks]

b.

## Examiners report

Many candidates readily earned marks in part (a). Some interpreted $${\log _2}40 – {\log _2}5$$ to mean $$\frac{{{{\log }_2}40}}{{{{\log }_2}5}}$$ , an error which led to no further marks. Others left the answer as $${\log _2}5$$ where an integer answer is expected.

a.

Part (b) proved challenging for most candidates, with few recognizing that changing $$8$$ to base $$2$$ is a helpful move. Some made it as far as $${2^{3{{\log }_2}5}}$$ yet could not make that final leap to an integer.

b.