# IB DP Maths Topic 1.1 :Exponents and logarithms HL Paper 1

## Question

Solve the equation $$2 – {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x$$ .

## Markscheme

$${\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}$$     M1M1A1

Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.

$$x + 7 = 18x$$     M1

$$x = \frac{7}{{17}}$$     A1

[5 marks]

## Examiners report

Some good solutions to this question and few candidates failed to earn marks on the question. Many were able to change the base of the logs, and many were able to deal with the 2, but of those who managed both, poor algebraic skills were often evident. Many students attempted to change the base into base 10, resulting in some complicated algebra, few of which managed to complete successfully.

## Question

Solve the equation $${8^{x – 1}} = {6^{3x}}$$. Express your answer in terms of $$\ln 2$$ and $$\ln 3$$.

## Markscheme

METHOD 1

$${2^{3(x – 1)}} = {(2 \times 3)^{3x}}$$     M1

Note:     Award M1 for writing in terms of 2 and 3.

$${2^{3x}} \times {2^{ – 3}} = {2^{3x}} \times {3^{3x}}$$

$${2^{ – 3}} = {3^{3x}}$$     A1

$$\ln \left( {{2^{ – 3}}} \right) = \ln \left( {{3^{3x}}} \right)$$     (M1)

$$– 3\ln 2 = 3x\ln 3$$     A1

$$x = – \frac{{\ln 2}}{{\ln 3}}$$     A1

METHOD 2

$$\ln {8^{x – 1}} = \ln {6^{3x}}$$     (M1)

$$(x – 1)\ln {2^3} = 3x\ln (2 \times 3)$$     M1A1

$$3x\ln 2 – 3\ln 2 = 3x\ln 2 + 3x\ln 3$$     A1

$$x = – \frac{{\ln 2}}{{\ln 3}}$$     A1

METHOD 3

$$\ln {8^{x – 1}} = \ln {6^{3x}}$$     (M1)

$$(x – 1)\ln 8 = 3x\ln 6$$     A1

$$x = \frac{{\ln 8}}{{\ln 8 – 3\ln 6}}$$     A1

$$x = \frac{{3\ln 2}}{{\ln \left( {\frac{{{2^3}}}{{{6^3}}}} \right)}}$$     M1

$$x = – \frac{{\ln 2}}{{\ln 3}}$$     A1

[5 marks]

[N/A]

## Question

Let $$\{ {u_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }$$, be an arithmetic sequence with first term equal to $$a$$ and common difference of $$d$$, where $$d \ne 0$$. Let another sequence $$\{ {v_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }$$, be defined by $${v_n} = {2^{{u_n}}}$$.

(i)     Show that $$\frac{{{v_{n + 1}}}}{{{v_n}}}$$ is a constant.

(ii)     Write down the first term of the sequence $$\{ {v_n}\}$$.

(iii)     Write down a formula for $${v_n}$$ in terms of $$a$$, $$d$$ and $$n$$.

[4]
a.

Let $${S_n}$$ be the sum of the first $$n$$ terms of the sequence $$\{ {v_n}\}$$.

(i)     Find $${S_n}$$, in terms of $$a$$, $$d$$ and $$n$$.

(ii)     Find the values of $$d$$ for which $$\sum\limits_{i = 1}^\infty {{v_i}}$$ exists.

You are now told that $$\sum\limits_{i = 1}^\infty {{v_i}}$$ does exist and is denoted by $${S_\infty }$$.

(iii)     Write down $${S_\infty }$$ in terms of $$a$$ and $$d$$ .

(iv)     Given that $${S_\infty } = {2^{a + 1}}$$ find the value of $$d$$ .

[8]
b.

Let $$\{ {w_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }$$, be a geometric sequence with first term equal to $$p$$ and common ratio $$q$$, where $$p$$ and $$q$$ are both greater than zero. Let another sequence $$\{ {z_n}\}$$ be defined by $${z_n} = \ln {w_n}$$.

Find $$\sum\limits_{i = 1}^n {{z_i}}$$ giving your answer in the form $$\ln k$$ with $$k$$ in terms of $$n$$, $$p$$ and $$q$$.

[6]
c.

## Markscheme

(i)     METHOD 1

$$\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}$$     M1

$$= {2^{{u_{n + 1}} – {u_n}}} = {2^d}$$     A1

METHOD 2

$$\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n – 1)d}}}}$$     M1

$$= {2^d}$$     A1

(ii)     $$= {2^a}$$     A1

Note:     Accept $$= {2^{{u_1}}}$$.

(iii)     EITHER

$${v_n}$$ is a GP with first term $${2^a}$$ and common ratio $${2^d}$$

$${v_n} = {2^a}{({2^d})^{(n – 1)}}$$

OR

$${u_n} = a + (n – 1)d$$ as it is an AP

THEN

$${v_n} = {2^a}^{ + (n – 1)d}$$     A1

[4 marks]

a.

(i)     $${S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} – 1} \right)}}{{{2^d} – 1}} = \frac{{{2^a}({2^{dn}} – 1)}}{{{2^d} – 1}}$$     M1A1

Note:     Accept either expression.

(ii)     for sum to infinity to exist need $$– 1 < {2^d} < 1$$     R1

$$\Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0$$     (M1)A1

Note:     Also allow graph of $${2^d}$$.

(iii)     $${S_\infty } = \frac{{{2^a}}}{{1 – {2^d}}}$$     A1

(iv)     $$\frac{{{2^a}}}{{1 – {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 – {2^d}}} = 2$$     M1

$$\Rightarrow 1 = 2 – {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1$$

$$\Rightarrow d = – 1$$     A1

[8 marks]

b.

METHOD 1

$${w_n} = p{q^{n – 1}},{\text{ }}{z_n} = \ln p{q^{n – 1}}$$     (A1)

$${z_n} = \ln p + (n – 1)\ln q$$     M1A1

$${z_{n + 1}} – {z_n} = (\ln p + n\ln q) – (\ln p + (n – 1)\ln q) = \ln q$$

which is a constant so this is an AP

(with first term $$\ln p$$ and common difference $$\ln q$$)

$$\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n – 1)\ln q} \right)}$$     M1

$$= n\left( {\ln p + \ln {q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right)$$     (M1)

$$= \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)$$     A1

METHOD 2

$$\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} + \ldots + \ln p{q^{n – 1}}}$$     (M1)A1

$$= \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 + \ldots + (n – 1)} \right)}}} \right)$$     (M1)A1

$$= \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)$$     (M1)A1

[6 marks]

Total [18 marks]

c.

## Examiners report

Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.

a.

Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving $$n$$.

b.

Not enough candidates realised that this was an AP.

c.

## Question

Find integer values of $$m$$ and $$n$$ for which

$m – n{\log _3}2 = 10{\log _9}6$

## Markscheme

METHOD 1

$$m – n{\log _3}2 = 10{\log _9}6$$

$$m – n{\log _3}2 = 5{\log _3}6$$ M1

$$m = {\log _3}\left( {{6^5}{2^n}} \right)$$ (M1)

$${3^m}{2^{ – n}} = {6^5} = {3^5} \times {2^5}$$ (M1)

$$m = 5,{\text{ }}n = – 5$$ A1

Note: First M1 is for any correct change of base, second M1 for writing as a single logarithm, third M1 is for writing 6 as $$2 \times 3$$.

METHOD 2

$$m – n{\log _3}2 = 10{\log _9}6$$

$$m – n{\log _3}2 = 5{\log _3}6$$ M1

$$m – n{\log _3}2 = 5{\log _3}3 + 5{\log _3}2$$ (M1)

$$m – n{\log _3}2 = 5 + 5{\log _3}2$$ (M1)

$$m = 5,{\text{ }}n = – 5$$ A1

Note: First M1 is for any correct change of base, second M1 for writing 6 as $$2 \times 3$$ and third M1 is for forming an expression without $${\log _3}3$$.

[4 marks]

## Examiners report

The first stage on this question was to change base, so each logarithm was written in the same base. Some candidates chose to move to base 10 or base e, rather than the more obvious base 3, but a few still successfully reached the correct answer having done this. A large majority though did not seem to know how to change the base of a logarithm.

Simplifying the expression further was a struggle for many candidates.

## Question

Solve the equation $${4^x} + {2^{x + 2}} = 3$$.

## Markscheme

attempt to form a quadratic in $${2^x}$$ M1

$${({2^x})^2} + 4 \bullet {2^x} – 3 = 0$$ A1

$${2^x} = \frac{{ – 4 \pm \sqrt {16 + 12} }}{2}{\text{ }}\left( { = – 2 \pm \sqrt 7 } \right)$$ M1

$${2^x} = – 2 + \sqrt 7 {\text{ }}\left( {{\text{as }} – 2 – \sqrt 7 < 0} \right)$$ R1

$$x = {\log _2}\left( { – 2 + \sqrt 7 } \right){\text{ }}\left( {x = \frac{{\ln \left( { – 2 + \sqrt 7 } \right)}}{{\ln 2}}} \right)$$ A1

Note: Award R0 A1 if final answer is $$x = {\log _2}\left( { – 2 + \sqrt 7 } \right)$$.

[5 marks]

[N/A]

## Question

Solve the equation $${\log _2}(x + 3) + {\log _2}(x – 3) = 4$$.

## Markscheme

$${\log _2}(x + 3) + {\log _2}(x – 3) = 4$$

$${\log _2}({x^2} – 9) = 4$$     (M1)

$${x^2} – 9 = {2^4}{\text{ }}( = 16)$$     M1A1

$${x^2} = 25$$

$$x = \pm 5$$     (A1)

$$x = 5$$     A1

[5 marks]

[N/A]

## Question

Solve $${\left( {{\text{ln}}\,x} \right)^2} – \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) < 2{\left( {{\text{ln}}\,2} \right)^2}$$.

## Markscheme

$${\left( {{\text{ln}}\,x} \right)^2} – \left( {{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x} \right) – 2{\left( {{\text{ln}}\,2} \right)^2}\left( { = 0} \right)$$

EITHER

$${\text{ln}}\,x = \frac{{{\text{ln}}\,2 \pm \sqrt {{{\left( {{\text{ln}}\,2} \right)}^2} + 8{{\left( {{\text{ln}}\,2} \right)}^2}} }}{2}$$ M1

$$= \frac{{{\text{ln}}\,2 \pm 3\,{\text{ln}}\,2}}{2}$$ A1

OR

$$\left( {{\text{ln}}\,x – 2\,{\text{ln}}\,2} \right)\left( {{\text{ln}}\,x + 2\,{\text{ln}}\,2} \right)\left( { = 0} \right)$$ M1A1

THEN

$${\text{ln}}\,x = 2\,{\text{ln}}\,2$$ or $$– {\text{ln}}\,2$$ A1

$$\Rightarrow x = 4$$ or $$x = \frac{1}{2}$$ (M1)A1

Note: (M1) is for an appropriate use of a log law in either case, dependent on the previous M1 being awarded, A1 for both correct answers.

solution is $$\frac{1}{2} < x < 4$$ A1

[6 marks]

[N/A]

## Question

It is given that $${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$.

Show that $${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$$ where $$r,\,x \in {\mathbb{R}^ + }$$.

[2]
a.

Express $$y$$ in terms of $$x$$. Give your answer in the form $$y = p{x^q}$$, where p , q are constants.

[5]
b.

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines $$x = 1$$ and $$x = \alpha$$ where $$\alpha > 1$$. The area of R is $$\sqrt 2$$.

Find the value of $$\alpha$$.

[5]
c.

## Markscheme

METHOD 1

$${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)$$ M1A1

$$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$$ AG

[2 marks]

METHOD 2

$${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}$$ M1

$$= \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}$$ A1

$$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$$ AG

[2 marks]

a.

METHOD 1

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y = – \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}$$

$${\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)$$ M1A1

$$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$$ A1

Note: For the final A mark, $$y$$ must be expressed in the form $$p{x^q}$$.

[5 marks]

METHOD 2

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$

$${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0$$ M1

$$\sqrt 2 xy = 1$$ A1

$$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$$ A1

Note: For the final A mark, $$y$$ must be expressed in the form $$p{x^q}$$.

[5 marks]

b.

the area of R is $$\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ – 1}}{\text{d}}x$$ M1

$$= \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha$$ A1

$$= \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha$$ A1

$$\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2$$ M1

$$\alpha = {{\text{e}}^2}$$ A1

Note: Only follow through from part (b) if $$y$$ is in the form $$y = p{x^q}$$

[5 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.

## Question

It is given that $${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$.

Show that $${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{2}{\text{lo}}{{\text{g}}_r}\,x$$ where $$r,\,x \in {\mathbb{R}^ + }$$.

[2]
a.

Express $$y$$ in terms of $$x$$. Give your answer in the form $$y = p{x^q}$$, where p , q are constants.

[5]
b.

The region R, is bounded by the graph of the function found in part (b), the x-axis, and the lines $$x = 1$$ and $$x = \alpha$$ where $$\alpha > 1$$. The area of R is $$\sqrt 2$$.

Find the value of $$\alpha$$.

[5]
c.

## Markscheme

METHOD 1

$${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{lo}}{{\text{g}}_r}\,{r^2}}}\left( { = \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{{{\text{2}}\,{\text{lo}}{{\text{g}}_r}\,r}}} \right)$$ M1A1

$$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$$ AG

[2 marks]

METHOD 2

$${\text{lo}}{{\text{g}}_{{r^2}}}x = \frac{1}{{{\text{lo}}{{\text{g}}_x}\,{r^2}}}$$ M1

$$= \frac{1}{{2\,{\text{lo}}{{\text{g}}_x}\,r}}$$ A1

$$= \frac{{{\text{lo}}{{\text{g}}_r}\,x}}{2}$$ AG

[2 marks]

a.

METHOD 1

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,2{x^2} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y = – \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2{x^2}$$

$${\text{lo}}{{\text{g}}_2}\,y = {\text{lo}}{{\text{g}}_2}\left( {\frac{1}{{\sqrt {2x} }}} \right)$$ M1A1

$$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$$ A1

Note: For the final A mark, $$y$$ must be expressed in the form $$p{x^q}$$.

[5 marks]

METHOD 2

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_4}\,x + {\text{lo}}{{\text{g}}_4}\,2x = 0$$

$${\text{lo}}{{\text{g}}_2}\,y + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,x + \frac{1}{2}{\text{lo}}{{\text{g}}_2}\,2x = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,y + {\text{lo}}{{\text{g}}_2}\,{x^{\frac{1}{2}}} + {\text{lo}}{{\text{g}}_2}\,{\left( {2x} \right)^{\frac{1}{2}}} = 0$$ M1

$${\text{lo}}{{\text{g}}_2}\,\left( {\sqrt 2 xy} \right) = 0$$ M1

$$\sqrt 2 xy = 1$$ A1

$$y = \frac{1}{{\sqrt 2 }}{x^{ – 1}}$$ A1

Note: For the final A mark, $$y$$ must be expressed in the form $$p{x^q}$$.

[5 marks]

b.

the area of R is $$\int\limits_1^\alpha {\frac{1}{{\sqrt 2 }}} {x^{ – 1}}{\text{d}}x$$ M1

$$= \left[ {\frac{1}{{\sqrt 2 }}{\text{ln}}\,x} \right]_1^\alpha$$ A1

$$= \frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha$$ A1

$$\frac{1}{{\sqrt 2 }}{\text{ln}}\,\alpha = \sqrt 2$$ M1

$$\alpha = {{\text{e}}^2}$$ A1

Note: Only follow through from part (b) if $$y$$ is in the form $$y = p{x^q}$$

[5 marks]

c.

[N/A]

a.

[N/A]

b.

[N/A]

c.
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