**Soalan 15 (4 markah):**

$\begin{array}{l}\left(\text{a}\right)\text{Diberi}P={\mathrm{log}}_{a}Q\text{,nyatakan}\\ \text{syarat-syaratbagi}a\text{.}\\ \left(\text{b}\right)\text{Diberi}{\mathrm{log}}_{3}y=\frac{2}{{\mathrm{log}}_{xy}3}\text{,ungkapkan}\\ y\text{dalamsebutan}x\text{.}\end{array}$

*Penyelesaian*:**(a)**

a > 0, a ≠ 1

**(b)**

$\begin{array}{l}{\mathrm{log}}_{3}y=\frac{2}{{\mathrm{log}}_{xy}3}\\ \frac{{\mathrm{log}}_{xy}y}{{\mathrm{log}}_{xy}3}=\frac{2}{{\mathrm{log}}_{xy}3}\\ {\mathrm{log}}_{xy}y=2\\ y={\left(xy\right)}^{2}\\ y={x}^{2}{y}^{2}\\ \frac{1}{{x}^{2}}=\frac{{y}^{2}}{y}\\ y=\frac{1}{{x}^{2}}\end{array}$

**Soalan 16 (2 markah):**

Diberi 2

*+ 2*

^{p}*= 2*

^{p}*. Ungkapkan*

^{k}*p*dalam sebutan

*k*.

*Penyelesaian*:$\begin{array}{l}{2}^{p}+{2}^{p}={2}^{k}\\ 2\left({2}^{p}\right)={2}^{k}\\ {2}^{p}=\frac{{2}^{k}}{{2}^{1}}\\ {2}^{p}={2}^{k-1}\\ p=k-1\end{array}$

**Soalan 17 (3 markah):**

$\begin{array}{l}\text{Diberi}\frac{{25}^{h+3}}{{125}^{p-1}}\text{=1,ungkapkan}p\text{dalam}\\ \text{sebutan}h\text{.}\end{array}$

*Penyelesaian*:$\begin{array}{l}\frac{{25}^{h+3}}{{125}^{p-1}}=1\\ {25}^{h+3}={125}^{p-1}\\ {\left({5}^{2}\right)}^{h+3}={\left({5}^{3}\right)}^{p-1}\\ {5}^{2h+6}={5}^{3p-3}\\ 2h+6=3p-3\\ 3p=2h+9\\ p=\frac{2h+9}{3}\end{array}$

**Soalan 18 (3 markah):**

$\begin{array}{l}\text{Selesaikanpersamaan:}\\ {\mathrm{log}}_{m}324-{\mathrm{log}}_{\sqrt{m}}2m=2\end{array}$

*Penyelesaian*:$\begin{array}{l}{\mathrm{log}}_{m}324-{\mathrm{log}}_{\sqrt{m}}2m=2\\ {\mathrm{log}}_{m}324-\frac{{\mathrm{log}}_{m}2m}{{\mathrm{log}}_{m}{m}^{\frac{1}{2}}}=2\\ {\mathrm{log}}_{m}324-2\left(\frac{{\mathrm{log}}_{m}2m}{{\mathrm{log}}_{m}m}\right)=2\\ {\mathrm{log}}_{m}324-2{\mathrm{log}}_{m}2m=2\\ {\mathrm{log}}_{m}324-{\mathrm{log}}_{m}{\left(2m\right)}^{2}=lo{g}_{m}{m}^{2}\\ {\mathrm{log}}_{m}\left(\frac{324}{4{m}^{2}}\right)=lo{g}_{m}{m}^{2}\\ \frac{324}{4{m}^{2}}={m}^{2}\\ 4{m}^{4}=324\\ {m}^{4}=81\\ m=\pm 3\left(-3\text{ditolak}\right)\end{array}$