 # 5.5.3 Indices and Logarithms, Short Questions (Question 11 – 13)

Question 11
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)

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
Solve the equation, $\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)$

Solution:
$\begin{array}{l}\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)\\ 2={\mathrm{log}}_{5}2.{\mathrm{log}}_{2}\left(2-x\right)\\ 2=\frac{1}{{\mathrm{log}}_{2}5}.{\mathrm{log}}_{2}\left(2-x\right)\\ 2{\mathrm{log}}_{2}5={\mathrm{log}}_{2}\left(2-x\right)\\ {\mathrm{log}}_{2}{5}^{2}={\mathrm{log}}_{2}\left(2-x\right)\\ 25=2-x\\ x=-23\end{array}$