IBDP Maths analysis and approaches Topic: SL 1.7: Laws of exponents; laws of logarithms HL Paper 1

Question

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

▶️Answer/Explanation

\({2^{2x – 2}} = {2^x} + 8\)     (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

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

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]

Question

Find integer values of \(m\) and \(n\) for which \[m – n{\log _3}2 = 10{\log _9}6\]

▶️Answer/Explanation

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]

Question

Solve the equation \({4^x} + {2^{x + 2}} = 3\).

▶️Answer/Explanation

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]

Scroll to Top