Question
a.Find \({\log _2}32\) .[1]
b.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]
Answer/Explanation
Markscheme
5 A1 N1
[1 mark]
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]
Question
Let \(f(x) = k{\log _2}x\) .
a.Given that \({f^{ – 1}}(1) = 8\) , find the value of \(k\) .[3]
b.Find \({f^{ – 1}}\left( {\frac{2}{3}} \right)\) .[4]
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]
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]
Question
Let \(f(x) = 3\ln x\) and \(g(x) = \ln 5{x^3}\) .
a. Express \(g(x)\) in the form \(f(x) + \ln a\) , where \(a \in {{\mathbb{Z}}^ + }\) .[4]
b.The graph of g is a transformation of the graph of f . Give a full geometric description of this transformation.[3]
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]
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]
Question
Let \({\log _3}p = 6\) and \({\log _3}q = 7\) .
a.Find \({\log _3}{p^2}\) .[2]
b. Find \({\log _3}\left( {\frac{p}{q}} \right)\) .[2]
c.Find \({\log _3}(9p)\) .[3]
Answer/Explanation
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]
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]
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]
Question
Solve the equations
(a) \(log_2(x+14)-2log_2x=2\)
(b) \(log_4(x+14)-log_2x=1\) by using change of base on \(log_4(x+14)\)
(c) \(log_2(x+14)=2+log_{\sqrt{2}}x\) by using change of base on \(log_{\sqrt{2}}x\)
Answer/Explanation
Ans
(a) x = 2 (b) x = 2 (c) x = 2 ( it is the same equation in all 3 cases )