Bab 5 Indeks dan Logaritma

Soalan 15 (4 markah):
$\begin{array}{l}\left(\text{a}\right)\text{ Diberi }P={\log }_{a}Q\text{, nyatakan }\\ \text{syarat-syarat bagi }a\text{.}\\ \left(\text{b}\right)\text{ Diberi }{\log }_{3}y=\frac{2}{{\log }_{xy}3}\text{, ungkapkan}\\ y\text{ dalam sebutan }x\text{.}\end{array}$

Penyelesaian:
(a)
a > 0, a ≠ 1

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

Soalan 16 (2 markah):
Diberi 2^p + 2^p = 2^k. Ungkapkan 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):
$\begin{array}{l}\text{Selesaikan persamaan:}\\ {\log }_{m}324-{\log }_{\sqrt{m}}2m=2\end{array}$

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