**5.5 Indeks dan Logaritma, SPM Praktis (Soalan Pendek)**

**Soalan 5:**

Selesaikan persamaan, ${\mathrm{log}}_{9}\left(x-2\right)={\mathrm{log}}_{3}2$

*Penyelesaian:*

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

Selesaikan persamaan, ${\mathrm{log}}_{9}\left(2x+12\right)={\mathrm{log}}_{3}\left(x+2\right)$

*Penyelesaian:*

$\begin{array}{l}{\mathrm{log}}_{9}\left(2x+12\right)={\mathrm{log}}_{3}\left(x+2\right)\\ \frac{{\mathrm{log}}_{3}\left(2x+12\right)}{{\mathrm{log}}_{3}9}={\mathrm{log}}_{3}\left(x+2\right)\\ {\mathrm{log}}_{3}\left(2x+12\right)=2{\mathrm{log}}_{3}\left(x+2\right)\\ {\mathrm{log}}_{3}\left(2x+12\right)={\mathrm{log}}_{3}{\left(x+2\right)}^{2}\\ 2x+12={x}^{2}+4x+4\\ {x}^{2}+2x-8=0\\ \left(x+4\right)\left(x-2\right)=0\\ x=-4\text{(tidakditerima)}\\ x=2\end{array}$
**Soalan 7:**

Selesaikan persamaan, ${\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3$

*Penyelesaian:*

$\begin{array}{l}{\mathrm{log}}_{4}x=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{{\mathrm{log}}_{2}4}=\frac{3}{2}{\mathrm{log}}_{2}3\\ \frac{{\mathrm{log}}_{2}x}{2}=\frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=2\times \frac{3}{2}{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x=3{\mathrm{log}}_{2}3\\ {\mathrm{log}}_{2}x={\mathrm{log}}_{2}{3}^{3}\\ x=27\end{array}$
**Soalan 8:**

Selesaikan persamaan, $\frac{2}{{\mathrm{log}}_{5}2}={\mathrm{log}}_{2}\left(2-x\right)$

*Penyelesaian:*

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