 # 5.5.2 Indices and Logarithms, Short Questions (Question 9 – 14)

Question 9
Solve the equation,  ${\mathrm{log}}_{2}4x=1-{\mathrm{log}}_{4}x$

Solution:
$\begin{array}{l}{\mathrm{log}}_{2}4x=1-{\mathrm{log}}_{4}x\\ {\mathrm{log}}_{2}4x=1-\frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}\\ {\mathrm{log}}_{2}4x=1-\frac{{\mathrm{log}}_{2}x}{2}\\ 2{\mathrm{log}}_{2}4x=2-{\mathrm{log}}_{2}x\\ {\mathrm{log}}_{2}16{x}^{2}={\mathrm{log}}_{2}4-{\mathrm{log}}_{2}x\\ {\mathrm{log}}_{2}16{x}^{2}={\mathrm{log}}_{2}\frac{4}{x}\\ 16{x}^{2}=\frac{4}{x}\\ {x}^{3}=\frac{4}{16}=\frac{1}{4}\\ x={\left(\frac{1}{4}\right)}^{\frac{1}{3}}=0.62996\end{array}$

Question 10
Solve the equation,  ${\mathrm{log}}_{4}x=25{\mathrm{log}}_{x}4$

Solution:

Question 11
Solve the equation,  $2{\mathrm{log}}_{x}5+{\mathrm{log}}_{5}x=\mathrm{lg}1000$

Solution:

Question 12
Solve the equation,  ${\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{4}16x=6$

Solution:
$\begin{array}{l}{\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{4}16x=6\\ {\mathrm{log}}_{2}5\sqrt{x}+\frac{{\mathrm{log}}_{2}16x}{{\mathrm{log}}_{2}4}=6\\ {\mathrm{log}}_{2}5\sqrt{x}+\frac{{\mathrm{log}}_{2}16x}{2}=6\\ 2{\mathrm{log}}_{2}5\sqrt{x}+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}{\left(5\sqrt{x}\right)}^{2}+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}\left(25x\right)+{\mathrm{log}}_{2}16x=12\\ {\mathrm{log}}_{2}\left(25x\right)\left(16x\right)=12\\ {\mathrm{log}}_{2}400{x}^{2}=12\\ 400{x}^{2}={2}^{12}\\ {x}^{2}=10.24\\ x=3.2\end{array}$

Question 13
Given that 2 log2 (xy) = 3 + log2x + log2 y
Prove that x2 + y2– 10xy = 0.

Solution:
2 log2 (xy) = 3 + log2x + log2 y
log2 (xy)2 = log2 8 + log2 x + log2y
log2 (xy)2 = log2 8xy
(xy)2 = 8xy
x2– 2xy + y2 = 8xy
x2 + y2 – 10xy = 0 (proven)