# Indices (CIE)

$\def\D{\displaystyle}$
1 (CIE 2012, s, paper 12, question 2)
Using the substitution $\D u = 2^x,$ find the values of $\D x$ such that $\D 2^{2x+2} = 5 (2^x) - 1 .$ [5]

2 (CIE 2012, s, paper 21, question 5)
(a) Solve the equation $\D 3^{2x} = 1000,$ giving your answer to 2 decimal places. [2]
(b) Solve the equation $\D \frac{36^{2y-5}}{6^{3y}}=\frac{6^{2y-1}}{216^{y+6}}.$ [4]

3 (CIE 2012, w, paper 11, question 4)
Using the substitution $\D u = 5^x,$ or otherwise, solve
$\D 5^{2x+1} = 7(5^x) - 2.$ [5]

4 (CIE 2012, w, paper 22, question 6)
(i) Given that $\D \frac{2^{x-3}}{8^{2y-3}}=16^{x-y},$  show that $\D 3x + 2y = 6.$ [2]
(ii) Given also that $\D \frac{5^y}{125^{x-2}}=25,$ find the value of $\D x$ and of $\D y.$ [4]

5 (CIE 2012, w, paper 23, question 11)
(a) Solve $\D \left( 2^{x-2}\right)^\frac{1}{2}=100,$  giving your answer to 1 decimal place. [3]
(b) Solve $\D \log_y 2 = 3 - \log_y 256.$ [3]
(c) Solve $\D \frac{6^{5z-2}}{36^z}= \frac{216^{z-1}}{36^{3-z}}.$

6 (CIE 2013, s, paper 22, question 2)
(a) Solve the equation $\D 3^{p+1} = 0.7 ,$ giving your answer to 2 decimal places. [3]
(b) Express $\D \frac{y\times (4x^3)^2}{\sqrt{8y^3}}$  in the form $\D 2^a \times x^b \times y^c,$ where $\D a, b$ and $\D c$ are constants. [3]

7 (CIE 2013, w, paper 21, question 5)
Solve the simultaneous equations
$\D \begin{array}{rcl} \frac{4^x}{256^y}&=&1024,\\ 3^{2x}\times 9^y&=&243. \end{array}$ [5]

8 (CIE 2014, s, paper 12, question 5)
(i) Given that $\D 2^{5x}\times 4^y= \frac{1}{8},$  show that $\D 5x + 2y = -3.$ [3]
(ii) Solve the simultaneous equations $\D 2^{5x}\times 4^y= \frac{1}{8}$ and $\D 7^x\times 49^{2y}=1.$ [4]

9 (CIE 2014, s, paper 13, question 2)
Given that $\D 2^{4x}\times 4^y\times 8^{x-y}=1$ and $\D 3^{x+y}= \frac{1}{3},$  find the value of $\D x$ and of $\D y.$ [4]

10 (CIE 2014, s, paper 21, question 11)
(a) Solve $\D 2^{x^2-5x}=\frac{1}{64}.$ [4]
(b) By changing the base of $\D \log_{2a} 4 ,$ express $\D (\log_{2a} 4)(1+\log_a 2)$ as a single logarithm to base $\D a.$ [4]

11 (CIE 2014, w, paper 11, question 4)
(i) Using the substitution $\D y = 5^x ,$ show that the equation $\D 5^{2x+1}-5^{x+1}+2=2(5^x)$ can be written in the form $\D ay^2+by+2=0,$ where $\D a$ and $\D b$ are constants to be found. [2]
(ii) Hence solve the equation $\D 5^{2x+1}-5^{x+1}+2=2(5^x).$ [4]

12 (CIE 2014, w, paper 13, question 10)
Solve the following simultaneous equations.
$\D \begin{array}{rcl} \frac{5^x}{25^{3y-2}}&=&1\\ \frac{3^x}{27^{y-1}}&=&81 \end{array}$ [5]

1. $\D x = 0,-2$
2. (a) $\D 3.14$
(b) $\D y = -4.5$
3. $\D x = 0;-0.569$
4. (ii) $\D x = 14/9; y = 2/3$
5. (a) $\D x = 15.3$
(b) $\D y = 8$
(c) $\D z = 3.5$
6. (a) $\D p = -1.32$
(b) $\D a = 5/2; b = 6; c = -1/2$
7. $\D x = 3; y = -0.5$
8. $\D x = -2/3; y = 1/6$
9. $\D x = -1/8; y = -7/8$
10. (a) $\D x = 2; 3$
(b) $\D \log_a 4$
11. (i) $\D 5y^2 - 7y + 2 = 0$
(ii) $\D x = 0; y = 1; x = -0.569; y =2/5$
12. (a) $\D x = 6; y = 5/3$