Saturday, December 22, 2018

Logarithms (CIE, Myanmar Grade 9)

$\def\D{\displaystyle}$
1 (CIE 2012, s, paper 11, question 8)
(a) Find the value of $\D x$ for which $\D 2\lg x - \lg(5x + 60) = 1 .$ [5]
(b) Solve $\D \log_5 y = 4\log_y 5 .$ [4]

2 (CIE 2012, w, paper 11, question 3)
Given that $\D p = \log_q 32,$ express, in terms of $\D p,$
(i) $\D \log_q 4,$ [2]
(ii) $\D \log_q 16q.$ [2]

3 (CIE 2012, w, paper 12, question 4)
Given that $\D \log_a pq = 9$ and $\D \log_a p^2q = 15,$ find the value of
(i) $\D \log_a p$ and of $\D \log_a q,$ [4]
(ii) $\D \log_p a + \log_q a.$ [2]

4 (CIE 2013, s, paper 11, question 4)
(i) Given that $\D \log_4 x=\frac{1}{2},$  find the value of $\D x.$ [1]
(ii) Solve $\D 2\log_4y- \log_4 (5y-12)=\frac{1}{2}.$ [4]

5 (CIE 2013, w, paper 13, question 2)
Solve $\D 2 \lg y - \lg(5y+60)=1.$ [5]

6 (CIE 2013, w, paper 21, question 4)
Given that $\D \log_p X= 5$ and $\D \log_p Y= 2,$ find
(i) $\D \log_p X^2,$ [1]
(ii) $\D \log_p\frac{1}{X},$  [1]
(iii) $\D \log_{XY} p .$ [2]

7 (CIE 2014, s, paper 22, question 6)
(a) (i) State the value of $\D u$ for which $\D \lg u = 0.$ [1]
(ii) Hence solve $\lg |2x + 3| = 0.$ [2]
(b) Express $\D 2 \log_315- (\log_a5) (\log_3a),$ where $\D a > 1,$ as a single logarithm to base 3. [4]

8 (CIE 2014, s, paper 23, question 2)
Using the substitution $\D u= \log_3 x,$ solve, for $\D x,$ the equation $\D \log_3x -12 \log_x3= 4 .$ [5]

9 (CIE 2014, w, paper 13, question 7)
Solve the equation $\D 1+ 2 \log_5 x= \log_5(18x-9).$ [5]

10 (CIE 2014, w, paper 21, question 3)
Solve the following simultaneous equations.
$\D \begin{array}{rcl}
\log_2(x+3)&=&2+\log_2y\\
\log_2(x+y)&=&3
\end{array}$ [5]

Answers
1. (a) $\D x = 60$
(b) $\D y = 25,\frac{1}{25}$
2. (i) $\D 2p/5$
(ii) $\D 1 + 4p/5$
3. (i) $\D \log_a p = 6,\log_a q = 3$
(ii) $\D 0.5$
4. (i) $\D 2$
(ii) $\D y = 4, 6$
5. $\D 60$
6. $\D 10;-5;1/7$
7. (a) $\D 1;x = -1,-2$
(b) $\log_3 45$
8. $\D 729; 1/9$
9. $\D x = 3/5; 3$
10. $\D x = 5.8; y = 2.2$

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