Home / IB DP Math AA Topic: SL 1.5: Laws of exponents with integer; laws of logarithms base 10 and e SL Paper 1

IB DP Math AA Topic: SL 1.5: Laws of exponents with integer; laws of logarithms base 10 and e SL Paper 1

Question

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

Answer/Explanation

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] 

Question

The first terms of an arithmetic sequence are \(\frac{1}{{{{\log }_2}x}},{\text{ }}\frac{1}{{{{\log }_8}x}},{\text{ }}\frac{1}{{{{\log }_{32}}x}},{\text{ }}\frac{1}{{{{\log }_{128}}x}},{\text{ }} \ldots \)

Find x if the sum of the first 20 terms of the sequence is equal to 100.

Answer/Explanation

Markscheme

METHOD 1

\(d = \frac{1}{{{{\log }_8}x}} – \frac{1}{{{{\log }_2}x}}\)     (M1)

\( = \frac{{{{\log }_2}8}}{{{{\log }_2}x}} – \frac{1}{{{{\log }_2}x}}\)     (M1)

Note: Award this M1 for a correct change of base anywhere in the question.

 

\( = \frac{2}{{{{\log }_2}x}}\)     (A1)

\(\frac{{20}}{2}\left( {2 \times \frac{1}{{{{\log }_2}x}} + 19 \times \frac{2}{{{{\log }_2}x}}} \right)\)     M1

\( = \frac{{400}}{{{{\log }_2}x}}\)     (A1)

\(100 = \frac{{400}}{{{{\log }_2}x}}\)

\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\)     A1

 

METHOD 2

\({20^{{\text{th}}}}{\text{ term}} = \frac{1}{{{{\log }_{{2^{39}}}}x}}\)     A1

\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_{{2^{39}}}}x}}} \right)\)     M1

\(100 = \frac{{20}}{2}\left( {\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}{2^{39}}}}{{{{\log }_2}x}}} \right)\)     M1(A1)

Note: Award this M1 for a correct change of base anywhere in the question.

 

\(100 = \frac{{400}}{{{{\log }_2}x}}\)     (A1)

\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\)     A1

 

METHOD 3

\(\frac{1}{{{{\log }_2}x}} + \frac{1}{{{{\log }_8}x}} + \frac{1}{{{{\log }_{32}}x}} + \frac{1}{{{{\log }_{128}}x}} +  \ldots \)

\(\frac{1}{{{{\log }_2}x}} + \frac{{{{\log }_2}8}}{{{{\log }_2}x}} + \frac{{{{\log }_2}32}}{{{{\log }_2}x}} + \frac{{{{\log }_2}128}}{{{{\log }_2}x}} +  \ldots \)     (M1)(A1)

Note: Award this M1 for a correct change of base anywhere in the question.

 

\( = \frac{1}{{{{\log }_2}x}}(1 + 3 + 5 +  \ldots )\)     A1

\( = \frac{1}{{{{\log }_2}x}}\left( {\frac{{20}}{2}(2 + 38)} \right)\)     (M1)(A1)

\(100 = \frac{{400}}{{{{\log }_2}x}}\)

\({\log _2}x = 4 \Rightarrow x = {2^4} = 16\)     A1

 

[6 marks]

Question

Solve the equation \({4^{x – 1}} = {2^x} + 8\).

Answer/Explanation

Markscheme

\({2^{2x – 2}} = {2^x} + 8\)     (M1) 

\(\frac{1}{4}{2^{2x}} = {2^x} + 8\)     (A1)

\({2^{2x}} – 4 \times {2^x} – 32 = 0\)     A1

\(({2^x} – 8)({2^x} + 4) = 0\)     (M1)

\({2^x} = 8 \Rightarrow x = 3\)     A1

Notes: Do not award final A1 if more than 1 solution is given.

 [5 marks]

Question

Consider \(a = {\log _2}3 \times {\log _3}4 \times {\log _4}5 \times  \ldots  \times {\log _{31}}32\). Given that \(a \in \mathbb{Z}\), find the value of a.

Answer/Explanation

Markscheme

\(\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} \times  \ldots \times \frac{{\log 32}}{{\log 31}}\)     M1A1

\( = \frac{{\log 32}}{{\log 2}}\)     A1

\( = \frac{{5\log 2}}{{\log 2}}\)     (M1)

\( = 5\)     A1

hence \(a = 5\)

Note:     Accept the above if done in a specific base eg \({\log _2}x\).

[5 marks]

Question

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

Answer/Explanation

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]

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