 # 4.5.3 Indices and Logarithms, Short Questions (Question 8 – 10)

Question 8
Solve the equation, ${\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2$

Solution:
$\begin{array}{l}{\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2\\ \overline{){\mathrm{log}}_{a}b=\frac{{\mathrm{log}}_{c}b}{{\mathrm{log}}_{c}a}}⇒\frac{{\mathrm{log}}_{3}\left(x-2\right)}{{\mathrm{log}}_{3}9}={\mathrm{log}}_{3}2\\ \frac{{\mathrm{log}}_{3}\left(x-2\right)}{2}={\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)=2{\mathrm{log}}_{3}2\\ {\mathrm{log}}_{3}\left(x-2\right)={\mathrm{log}}_{3}{2}^{2}\\ x-2=4\\ x=6\end{array}$

Question 9
Solve the equation, ${\mathrm{log}}_{9}\left(2x+12\right)={\mathrm{log}}_{3}\left(x+2\right)$

Solution:

Question 10
Solve the equation, ${\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3$

Solution:
$\begin{array}{l}{\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{2}=\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=2×\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=3{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x={\mathrm{log}}_{2}{3}^{3}\\ x=27\end{array}$