Question
Let \(f(x) = {{\rm{e}}^{x + 3}}\) .
(i) Show that \({f^{ – 1}}(x) = \ln x – 3\) .
(ii) Write down the domain of \({f^{ – 1}}\) .
Solve the equation \({f^{ – 1}}(x) = \ln \frac{1}{x}\) .
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]
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]
Examiners report
Many candidates interchanged the \(x\) and \(y\) to find the inverse function, but very few could write down the correct domain of the inverse, often giving \(x \ge 0\) , \(x > 3\) and “all real numbers” as responses.
Where students attempted to solve the equation in (b), most treated \(\ln x – 3\) as \(\ln (x – 3)\) and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.
Question
Find the value of \({\log _2}40 – {\log _2}5\) .
Find the value of \({8^{{{\log }_2}5}}\) .
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]
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]
Examiners report
Many candidates readily earned marks in part (a). Some interpreted \({\log _2}40 – {\log _2}5\) to mean \(\frac{{{{\log }_2}40}}{{{{\log }_2}5}}\) , an error which led to no further marks. Others left the answer as \({\log _2}5\) where an integer answer is expected.
Part (b) proved challenging for most candidates, with few recognizing that changing \(8\) to base \(2\) is a helpful move. Some made it as far as \({2^{3{{\log }_2}5}}\) yet could not make that final leap to an integer.