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]