IB Math Analysis & Approaches Questionbank-Topic: SL 1.7 Laws of logarithms- SL Paper 1

Question

Let \(f(x) = {{\rm{e}}^{x + 3}}\) .

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

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

[3]
a.

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

[4]
b.
Answer/Explanation

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.

Question

Find \({\log _2}32\) .

[1]
a.

Given that \({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right)\) can be written as \(px + qy\) , find the value of p and of q.

[4]
b.
Answer/Explanation

Markscheme

5     A1     N1

[1 mark]

a.

METHOD 1

\({\log _2}\left( {\frac{{{{32}^x}}}{{{8^y}}}} \right) = {\log _2}{32^x} – {\log _2}{8^y}\)     (A1)

\( = x{\log _2}32 – y{\log _2}8\)     (A1)

\({\log _2}8 = 3\)     (A1)

\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) )     A1      N3 

METHOD 2

\(\frac{{{{32}^x}}}{{{8^y}}} = \frac{{{{({2^5})}^x}}}{{{{({2^3})}^y}}}\)     (A1) 

\( = \frac{{{2^5}^x}}{{{2^3}^y}}\)     (A1)

\( = {2^{5x – 3y}}\)     (A1)

\({\log _2}({2^{5x – 3y}}) = 5x – 3y\)

\(p = 5\) , \(q = – 3\) (accept \(5x – 3y\) )     A1      N3

[4 marks]

b.

Question

Let \(f(x) = k{\log _2}x\) .

Given that \({f^{ – 1}}(1) = 8\) , find the value of \(k\) .

[3]
a.

Find \({f^{ – 1}}\left( {\frac{2}{3}} \right)\) .

[4]
b.
Answer/Explanation

Markscheme

METHOD 1

recognizing that \(f(8) = 1\)     (M1)

e.g. \(1 = k{\log _2}8\)

recognizing that \({\log _2}8 = 3\)     (A1)

e.g. \(1 = 3k\)

\(k = \frac{1}{3}\)     A1     N2

METHOD 2

attempt to find the inverse of \(f(x) = k{\log _2}x\)     (M1)

e.g. \(x = k{\log _2}y\) , \(y = {2^{\frac{x}{k}}}\)

substituting 1 and 8     (M1)

e.g. \(1 = k{\log _2}8\) , \({2^{\frac{1}{k}}} = 8\)

\(k = \frac{1}{{{{\log }_2}8}}\) \(\left( {k = \frac{1}{3}} \right)\)     A1     N2

[3 marks]

a.

METHOD 1

recognizing that \(f(x) = \frac{2}{3}\)     (M1)

e.g. \(\frac{2}{3} = \frac{1}{3}{\log _2}x\)

\({\log _2}x = 2\)     (A1)

\({f^{ – 1}}\left( {\frac{2}{3}} \right) = 4\) (accept \(x = 4\))     A2     N3

METHOD 2

attempt to find inverse of \(f(x) = \frac{1}{3}{\log _2}x\)     (M1)

e.g. interchanging x and y , substituting \(k = \frac{1}{3}\) into \(y = {2^{\frac{x}{k}}}\)

correct inverse     (A1)

e.g. \({f^{ – 1}}(x) = {2^{3x}}\) , \({2^{3x}}\)

\({f^{ – 1}}\left( {\frac{2}{3}} \right) = 4\)     A2    N3

[4 marks]

b.

Question

Let \(f(x) = lo{g_3}\sqrt x \) , for \(x > 0\) .

Show that \({f^{ – 1}}(x) = {3^{2x}}\) .

[2]
a.

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

[1]
b.

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ – 1}} \circ g)(2)\) , giving your answer as an integer.

[4]
c.
Answer/Explanation

Markscheme

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ – 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

a.

\(y > 0\) , \({f^{ – 1}}(x) > 0\)     A1     N1

[1 mark]

b.

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ – 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

c.

Question

Solve \({\log _2}x + {\log _2}(x – 2) = 3\) , for \(x > 2\) .

Answer/Explanation

Markscheme

recognizing \(\log a + \log b = \log ab\) (seen anywhere)     (A1)

e.g. \({\log _2}(x(x – 2))\) , \({x^2} – 2x\)

recognizing \({\log _a}b = x \Leftrightarrow {a^x} = b\)     (A1)

e.g. \({2^3} = 8\)

correct simplification     A1

e.g. \(x(x – 2) = {2^3}\) , \({x^2} – 2x – 8\)

evidence of correct approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((x – 4)(x + 2)\) , \(\frac{{2 \pm \sqrt {36} }}{2}\)

\(x = 4\)     A2     N3

[7 marks]

Question

Let \(f(x) = 3\ln x\) and \(g(x) = \ln 5{x^3}\) .

Express \(g(x)\) in the form \(f(x) + \ln a\) , where \(a \in {{\mathbb{Z}}^ + }\) .

[4]
a.

The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.

[3]
b.
Answer/Explanation

Markscheme

attempt to apply rules of logarithms     (M1)

e.g. \(\ln {a^b} = b\ln a\) , \(\ln ab = \ln a + \ln b\)

correct application of \(\ln {a^b} = b\ln a\) (seen anywhere)     A1

e.g. \(3\ln x = \ln {x^3}\)

correct application of \(\ln ab = \ln a + \ln b\) (seen anywhere)     A1

e.g. \(\ln 5{x^3} = \ln 5 + \ln {x^3}\)

so \(\ln 5{x^3} = \ln 5 + 3\ln x\)

\(g(x) = f(x) + \ln 5\) (accept \(g(x) = 3\ln x + \ln 5\) )     A1     N1

[4 marks]

a.

transformation with correct name, direction, and value     A3

e.g. translation by \(\left( {\begin{array}{*{20}{c}}
0\\
{\ln 5}
\end{array}} \right)\) , shift up by \(\ln 5\) , vertical translation of \(\ln 5\)

[3 marks]

b.

Question

Let \(f(x) = \frac{1}{4}{x^2} + 2\)  . The line L is the tangent to the curve of f at (4, 6) .

Let \(g(x) = \frac{{90}}{{3x + 4}}\) , for \(2 \le x \le 12\) . The following diagram shows the graph of g .

Find the equation of L .

[4]
a.

Find the area of the region enclosed by the curve of g , the x-axis, and the lines \(x = 2\) and \(x = 12\) . Give your answer in the form \(a\ln b\) , where \(a,b \in \mathbb{Z}\) .

[6]
b.

The graph of g is reflected in the x-axis to give the graph of h . The area of the region enclosed by the lines L , \(x = 2\) , \(x = 12\) and the x-axis is 120 \(120{\text{ c}}{{\text{m}}^2}\) .

Find the area enclosed by the lines L , \(x = 2\) , \(x = 12\) and the graph of h .

[3]
c.
Answer/Explanation

Markscheme

finding \(f'(x) = \frac{1}{2}x\)     A1

attempt to find \(f'(4)\)     (M1)

correct value \(f'(4) = 2\)     A1

correct equation in any form     A1     N2

e.g. \(y – 6 = 2(x – 4)\) , \(y = 2x – 2\)

[4 marks]

a.

\({\rm{area}} = \int_2^{12} {\frac{{90}}{{3x + 4}}} {\rm{d}}x\)

correct integral     A1A1

e.g. \(30\ln (3x + 4)\)

substituting limits and subtracting     (M1)

e.g. \(30\ln (3 \times 12 + 4) – 30\ln (3 \times 2 + 4)\) , \(30\ln 40 – 30\ln 10\)

correct working     (A1)

e.g. \(30(\ln 40 – \ln 10)\)

correct application of \(\ln b – \ln a\)     (A1)

e.g. \(30\ln \frac{{40}}{{10}}\)

\({\rm{area}} = 30\ln 4\)     A1     N4

[6 marks]

b.

valid approach     (M1)

e.g. sketch, area h = area g , 120 + their answer from (b)

\({\rm{area}} = 120 + 30\ln 4\)     A2     N3

[3 marks]

c.

Question

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

[3]
a.

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

[4]
b.
Answer/Explanation

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 \({\log _3}p = 6\) and \({\log _3}q = 7\) .

Find \({\log _3}{p^2}\) .

[2]
a.

Find \({\log _3}\left( {\frac{p}{q}} \right)\) .

[2]
b.

Find \({\log _3}(9p)\) .

[3]
c.
Answer/Explanation

Markscheme

METHOD 1

evidence of correct formula     (M1)

eg   \(\log {u^n} = n\log u\) , \(2{\log _3}p\)

\({\log _3}({p^2}) = 12\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\)     (M1)

eg \({\log _3}{({3^6})^2}\) , \(\log {3^{12}}\) , \(12{\log _3}3\)

\({\log _3}({p^2}) = 12\)     A1     N2

[2 marks]

a.

METHOD 1

evidence of correct formula     (M1)

eg   \(\log \left( {\frac{p}{q}} \right) = \log p – \log q\) , \(6 – 7\)

\({\log _3}\left( {\frac{p}{q}} \right) =  – 1\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\) and \(q = {3^7}\)     (M1)

eg   \({\log _3}\left( {\frac{{{3^6}}}{{{3^7}}}} \right)\) , \(\log {3^{ – 1}}\) , \( – {\log _3}3\)

\({\log _3}\left( {\frac{p}{q}} \right) =  – 1\)     A1     N2

[2 marks]

b.

METHOD 1

evidence of correct formula     (M1)

eg   \({\log _3}uv = {\log _3}u + {\log _3}v\) , \(\log 9 + \log p\)

\({\log _3}9 = 2\) (may be seen in expression)     A1

eg   \(2 + \log p\)

\({\log _3}(9p) = 8\)     A1     N2

METHOD 2

valid method using \(p = {3^6}\)     (M1)

eg   \({\log _3}(9 \times {3^6})\) , \({\log _3}({3^2} \times {3^6})\)

correct working     A1

eg   \({\log _3}9 + {\log _3}{3^6}\) , \({\log _3}{3^8}\)

\({\log _3}(9p) = 8\)     A1     N2

[3 marks]

Total [7 marks]

c.

Question

Find the value of each of the following, giving your answer as an integer.

\({\log _6}36\)

[2]
a.

\({\log _6}4 + {\log _6}9\)

[2]
b.

\({\log _6}2 – {\log _6}12\)

[3]
c.
Answer/Explanation

Markscheme

correct approach     (A1)

eg     \({6^x} = 36,{\text{ }}{6^2}\)

\(2\)      A1     N2

[2 marks]

a.

correct simplification     (A1)

eg     \({\log _6}36,{\text{ }}\log (4 \times 9)\)

\(2\)      A1      N2

[2 marks]

b.

correct simplification     (A1)

eg     \({\log _6}\frac{2}{{12}},{\text{ }}\log (2 \div 12)\)

correct working     (A1)

eg     \({\log _6}\frac{1}{6},{\text{ }}{6^{ – 1}} = \frac{1}{6}{,6^x} = \frac{1}{6}\)

\(-1\)     A1     N2

[3 marks]

c.

Question

Write the expression \(3\ln 2 – \ln 4\) in the form \(\ln k\), where \(k \in \mathbb{Z}\).

[3]
a.

Hence or otherwise, solve \(3\ln 2 – \ln 4 =  – \ln x\).

[3]
b.
Answer/Explanation

Markscheme

correct application of \(\ln {a^b} = b\ln a\) (seen anywhere)     (A1)

eg\(\;\;\;\ln 4 = 2\ln 2,{\text{ }}3\ln 2 = \ln {2^3},{\text{ }}3\log 2 = \log 8\)

correct working     (A1)

eg\(\;\;\;3\ln 2 – 2\ln 2,{\text{ }}\ln 8 – \ln 4\)

\(\ln 2\;\;\;{\text{(accept }}k = 2{\text{)}}\)     A1     N2

[3 marks]

a.

METHOD 1

attempt to substitute their answer into the equation     (M1)

eg\(\;\;\;\ln 2 =  – \ln x\)

correct application of a log rule     (A1)

eg\(\;\;\;\ln \frac{1}{x},{\text{ }}\ln \frac{1}{2} = \ln x,{\text{ }}\ln 2 + \ln x = \ln 2x\;\;\;( = 0)\)

\(x = \frac{1}{2}\)     A1     N2

METHOD 2

attempt to rearrange equation, with  \(3\ln 2\) written as \(\ln {2^3}\) or \(\ln 8\)     (M1)

eg\(\;\;\;\ln x = \ln 4 – \ln {2^3},{\text{ }}\ln 8 + \ln x = \ln 4,{\text{ }}\ln {2^3} = \ln 4 – \ln x\)

correct working applying \(\ln a \pm \ln b\)     (A1)

eg\(\;\;\;\frac{4}{8},{\text{ }}8x = 4,{\text{ }}\ln {2^3} = \ln \frac{4}{x}\)

\(x = \frac{1}{2}\)     A1     N2

[3 marks]

Total [6 marks]

b.

Question

Given that \({2^m} = 8\) and \({2^n} = 16\), write down the value of \(m\) and of \(n\).

[2]
a.

Hence or otherwise solve \({8^{2x + 1}} = {16^{2x – 3}}\).

[4]
b.
Answer/Explanation

Markscheme

\(m = 3,{\text{ }}n = 4\)     A1A1     N2

[2 marks]

a.

attempt to apply \({({2^a})^b} = {2^{ab}}\)     (M1)

eg\(\;\;\;6x + 3,{\text{ }}4(2x – 3)\)

equating their powers of \(2\) (seen anywhere)     M1

eg\(\;\;\;3(2x + 1) = 8x – 12\)

correct working     A1

eg\(\;\;\;8x – 12 = 6x + 3,{\text{ }}2x = 15\)

\(x = \frac{{15}}{2}\;\;\;(7.5)\)     A1     N2

[4 marks]

Total [6 marks]

b.

Question

An arithmetic sequence has the first term \(\ln a\) and a common difference \(\ln 3\).

The 13th term in the sequence is \(8\ln 9\). Find the value of \(a\).

Answer/Explanation

Markscheme

Note:     There are many approaches to this question, and the steps may be done in any order. There are 3 relationships they may need to apply at some stage, for the 3rd, 4th and 5th marks. These are

equating bases eg recognising 9 is \({{\text{3}}^2}\)

log rules: \(\ln b + \ln c = \ln (bc),{\text{ }}\ln b – \ln c = \ln \left( {\frac{b}{c}} \right)\),

exponent rule: \(\ln {b^n} = n\ln b\).

The exception to the FT rule applies here, so that if they demonstrate correct application of the 3 relationships, they may be awarded the A marks, even if they have made a previous error. However all applications of a relationship need to be correct. Once an error has been made, do not award A1FT for their final answer, even if it follows from their working.

Please check working and award marks in line with the markscheme.

correct substitution into \({u_{13}}\) formula     (A1)

eg\(\;\;\;\ln a + (13 – 1)\ln 3\)

set up equation for \({u_{13}}\) in any form (seen anywhere)     (M1)

eg\(\;\;\;\ln a + 12\ln 3 = 8\ln 9\)

correct application of relationships     (A1)(A1)(A1)

\(a = 81\)     A1     N3

[6 marks]

Examples of application of relationships

Example 1

correct application of exponent rule for logs     (A1)

eg\(\;\;\;\ln a + \ln {3^{12}} = \ln {9^8}\)

correct application of addition rule for logs     (A1)

eg\(\;\;\;\ln (a{3^{12}}) = \ln {9^8}\)

substituting for 9 or 3 in ln expression in equation     (A1)

eg\(\;\;\;\ln (a{3^{12}}) = \ln {3^{16}},{\text{ }}\ln (a{9^6}) = \ln {9^8}\)

Example 2

recognising \(9 = {3^2}\)      (A1)

eg\(\;\;\;\ln a + 12\ln 3 = 8\ln {3^2},{\text{ }}\ln a + 12\ln {9^{\frac{1}{2}}} = 8\ln 9\)

one correct application of exponent rule for logs relating \(\ln 9\) to \(\ln 3\)     (A1)

eg\(\;\;\;\ln a + 12\ln 3 = 16\ln 3,{\text{ }}\ln a + 6\ln 9 = 8\ln 9\)

another correct application of exponent rule for logs     (A1)

eg\(\;\;\;\ln a = \ln {3^4},{\text{ }}\ln a = \ln {9^2}\)

Question

Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).

The graph of \(f\) has a maximum point at P.

The \(y\)-coordinate of P is \(\ln 27\).

Find the \(x\)-coordinate of P.

[3]
a.

Find \(f(x)\), expressing your answer as a single logarithm.

[8]
b.

The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).

Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).

[[N/A]]
c.
Answer/Explanation

Markscheme

recognizing \(f'(x) = 0\)     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6 – 2x = 0\)

\(x = 3\)    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)

using substitution     (A1)

eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)

correct integral     A1

eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)

substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\))     (M1)

eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)

EITHER

\(c = \ln 3\)    (A1)

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\)     A1     N4

OR

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)

correct use of a log law     (A1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)

\(f(x) = \ln \left( {3(6x – {x^2})} \right)\)    A1     N4

[8 marks]

b.

\(a = 3\)    A1     N1

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)

correct use of log law     (A1)

eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)

\(b = 3\)    A1     N2

[4 marks]

c.

Question

Let \(x = \ln 3\) and \(y = \ln 5\). Write the following expressions in terms of \(x\) and \(y\).

\(\ln \left( {\frac{5}{3}} \right)\).

[2]
a.

\(\ln 45\).

[4]
b.
Answer/Explanation

Markscheme

correct approach     (A1)

eg\(\,\,\,\,\,\)\(\ln 5 – \ln 3\)

\(\ln \left( {\frac{5}{3}} \right) = y – x\)    A1     N2

[2 marks]

a.

recognizing factors of 45 (may be seen in log expansion)     (M1)

eg\(\,\,\,\,\,\)\(\ln (9 \times 5),{\text{ }}3 \times 3 \times 5,{\text{ }}\log {3^2} \times \log 5\)

correct application of \(\log (ab) = \log a + \log b\)     (A1)

eg\(\,\,\,\,\,\)\(\ln 9 + \ln 5,{\text{ }}\ln 3 + \ln 3 + \ln 5,{\text{ }}\ln {3^2} + \ln 5\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(2\ln 3 + \ln 5,{\text{ }}x + x + y\)

\(\ln 45 = 2x + y\)    A1     N3

[4 marks]

b

Question

The first two terms of an infinite geometric sequence, in order, are

\(2{\log _2}x,{\text{ }}{\log _2}x\), where \(x > 0\).

The first three terms of an arithmetic sequence, in order, are

\({\log _2}x,{\text{ }}{\log _2}\left( {\frac{x}{2}} \right),{\text{ }}{\log _2}\left( {\frac{x}{4}} \right)\), where \(x > 0\).

Let \({S_{12}}\) be the sum of the first 12 terms of the arithmetic sequence.

Find \(r\).

[2]
a.

Show that the sum of the infinite sequence is \(4{\log _2}x\).

[2]
b.

Find \(d\), giving your answer as an integer.

[4]
c.

Show that \({S_{12}} = 12{\log _2}x – 66\).

[2]
d.

Given that \({S_{12}}\) is equal to half the sum of the infinite geometric sequence, find \(x\), giving your answer in the form \({2^p}\), where \(p \in \mathbb{Q}\).

[3]
e.
Answer/Explanation

Markscheme

evidence of dividing terms (in any order)     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{\mu _2}}}{{{\mu _1}}},{\text{ }}\frac{{2{{\log }_2}x}}{{{{\log }_2}x}}\)

\(r = \frac{1}{2}\)    A1     N2

[2 marks]

a.

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{1 – \frac{1}{2}}}\)

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{2{{\log }_2}x}}{{\frac{1}{2}}}\)

\({S_\infty } = 4{\log _2}x\)     AG     N0

[2 marks]

b.

evidence of subtracting two terms (in any order)     (M1)

eg\(\,\,\,\,\,\)\({u_3} – {u_2},{\text{ }}{\log _2}x – {\log _2}\frac{x}{2}\)

correct application of the properties of logs     (A1)

eg\(\,\,\,\,\,\)\({\log _2}\left( {\frac{{\frac{x}{2}}}{x}} \right),{\text{ }}{\log _2}\left( {\frac{x}{2} \times \frac{1}{x}} \right),{\text{ }}({\log _2}x – {\log _2}2) – {\log _2}x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({\log _2}\frac{1}{2},{\text{ }} – {\log _2}2\)

\(d =  – 1\)    A1     N3

[4 marks]

c.

correct substitution into the formula for the sum of an arithmetic sequence     (A1)

eg\(\,\,\,\,\,\)\(\frac{{12}}{2}\left( {2{{\log }_2}x + (12 – 1)( – 1)} \right)\)

correct working     A1

eg\(\,\,\,\,\,\)\(6(2{\log _2}x – 11),{\text{ }}\frac{{12}}{2}(2{\log _2}x – 11)\)

\(12{\log _2}x – 66\)    AG     N0

[2 marks]

d.

correct equation     (A1)

eg\(\,\,\,\,\,\)\(12{\log _2}x – 66 = 2{\log _2}x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(10{\log _2}x = 66,{\text{ }}{\log _2}x = 6.6,{\text{ }}{2^{66}} = {x^{10}},{\text{ }}{\log _2}\left( {\frac{{{x^{12}}}}{{{x^2}}}} \right) = 66\)

\(x = {2^{6.6}}\) (accept \(p = \frac{{66}}{{10}}\))     A1     N2

[3 marks]

e.

Question

The first three terms of a geometric sequence are \(\ln {x^{16}}\), \(\ln {x^8}\), \(\ln {x^4}\), for \(x > 0\).

Find the common ratio.

[3]
a.

Solve \(\sum\limits_{k = 1}^\infty  {{2^{5 – k}}\ln x = 64} \).

[5]
b.
Answer/Explanation

Markscheme

correct use \(\log {x^n} = n\log x\)     A1

eg\(\,\,\,\,\,\)\(16\ln x\)

valid approach to find \(r\)     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{u_{n + 1}}}}{{{u_n}}},{\text{ }}\frac{{\ln {x^8}}}{{\ln {x^{16}}}},{\text{ }}\frac{{4\ln x}}{{8\ln x}},{\text{ }}\ln {x^4} = \ln {x^{16}} \times {r^2}\)

\(r = \frac{1}{2}\)     A1     N2

[3 marks]

a.

recognizing a sum (finite or infinite)     (M1)

eg\(\,\,\,\,\,\)\({2^4}\ln x + {2^3}\ln x,{\text{ }}\frac{a}{{1 – r}},{\text{ }}{S_\infty },{\text{ }}16\ln x +  \ldots \)

valid approach (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)recognizing GP is the same as part (a), using their \(r\) value from part (a), \(r = \frac{1}{2}\)

correct substitution into infinite sum (only if \(\left| r \right|\) is a constant and less than 1)     A1

eg\(\,\,\,\,\,\)\(\frac{{{2^4}\ln x}}{{1 – \frac{1}{2}}},{\text{ }}\frac{{\ln {x^{16}}}}{{\frac{1}{2}}},{\text{ }}32\ln x\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\ln x = 2\)

\(x = {{\text{e}}^2}\)     A1     N3

[5 marks]

b.

Question

An arithmetic sequence has \({u_1} = {\text{lo}}{{\text{g}}_c}\left( p \right)\) and \({u_2} = {\text{lo}}{{\text{g}}_c}\left( {pq} \right)\), where \(c > 1\) and \(p,\,\,q > 0\).

Show that \(d = {\text{lo}}{{\text{g}}_c}\left( q \right)\).

[2]
a.

Let \(p = {c^2}\) and \(q = {c^3}\). Find the value of \(\sum\limits_{n = 1}^{20} {{u_n}} \).

[6]
b.
Answer/Explanation

Markscheme

valid approach involving addition or subtraction       M1
eg  \({u_2} = {\text{lo}}{{\text{g}}_c}\,p + d,\,\,{u_1} – {u_2}\)

correct application of log law      A1
eg  \({\text{lo}}{{\text{g}}_c}\left( {pq} \right) = {\text{lo}}{{\text{g}}_c}\,p + {\text{lo}}{{\text{g}}_c}\,q,\,\,{\text{lo}}{{\text{g}}_c}\left( {\frac{{pq}}{p}} \right)\)

\(d = {\text{lo}}{{\text{g}}_c}\,q\)    AG N0

[2 marks]

a.

METHOD 1 (finding \({u_1}\) and d)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

attempt to find \({u_1}\) or d using \({\text{lo}}{{\text{g}}_c}\,{c^k} = k\)     (M1)
eg  \({\text{lo}}{{\text{g}}_c}\,c\), \({\text{3}}\,{\text{lo}}{{\text{g}}_c}\,c\), correct value of \({u_1}\) or d

\({u_1}\) = 2, d = 3 (seen anywhere)      (A1)(A1)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 2 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

\({\text{lo}}{{\text{g}}_c}\,{c^2} = 2,\,\,\,{\text{lo}}{{\text{g}}_c}\,{c^3} = 3\)  (seen anywhere)     (A1)(A1)

correct working      (A1)

eg  \({S_{20}} = \frac{{20}}{2}\left( {2 \times 2 + 19 \times 3} \right),\,\,{S_{20}} = \frac{{20}}{2}\left( {2 + 59} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

METHOD 3 (expressing S in terms of c)

recognizing \(\sum { = {S_{20}}} \) (seen anywhere)      (A1)

correct expression for S in terms of c      (A1)
eg  \(10\left( {2\,{\text{lo}}{{\text{g}}_c}\,{c^2} + 19\,{\text{lo}}{{\text{g}}_c}\,{c^3}} \right)\)

correct application of log law     (A1)
eg  \(2\,{\text{lo}}{{\text{g}}_c}\,{c^2} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^4},\,\,19\,{\text{lo}}{{\text{g}}_c}\,{c^3} = \,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}},\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^2}} \right)}^2} + \,\,{\text{lo}}{{\text{g}}_c}\,{{\left( {{c^3}} \right)}^{19}}} \right),\,\,10\,\left( {{\text{lo}}{{\text{g}}_c}\,{c^4} + \,{\text{lo}}{{\text{g}}_c}\,{c^{57}}} \right),\,\,10\left( {{\text{lo}}{{\text{g}}_c}\,{c^{61}}} \right)\)

correct application of definition of log      (A1)
eg  \({\text{lo}}{{\text{g}}_c}\,{c^{61}} = 61,\,\,{\text{lo}}{{\text{g}}_c}\,{c^4} = 4,\,\,{\text{lo}}{{\text{g}}_c}\,{c^{57}} = 57\)

correct working     (A1)
eg  \({S_{20}} = \frac{{20}}{2}\left( {4 + 57} \right),\,\,10\left( {61} \right)\)

\(\sum\limits_{n = 1}^{20} {{u_n}} \) = 610     A1 N2

[6 marks]

b.

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