**5.6 Indeks dan Logaritma, SPM Praktis (Soalan Panjang)**

**Soalan 3:**

Diberi bahawa

*p*= 3*dan*^{r}*q*= 3*, ungkapkan yang berikut dalam sebutan*^{t}*r*dan/ atau*t*.$\text{(a)}{\mathrm{log}}_{3}\frac{p{q}^{2}}{27},$

(b)
log

_{9}*p*– log_{27}*q*.

*Penyelesaian:***(a)**

Diberi

*p*= 3*, log*^{r}_{3}*p*=*r*^{}*q*= 3

*, log*

^{t}_{3}

*q*=

*t*

${\mathrm{log}}_{3}\frac{p{q}^{2}}{27}$

= log

_{3}*pq*^{2}– log_{3}27= log

_{3}*p*+ log_{3}*q*^{2}– log_{3}3^{3}=

*r*+ 2 log_{3}*q*– 3 log_{3}3=

*r*+ 2 log_{3}*q*– 3(1)^{}=

*r*+ 2*t*– 3^{}**(b)**

log

$\begin{array}{l}=\frac{{\mathrm{log}}_{3}p}{{\mathrm{log}}_{3}9}-\frac{{\mathrm{log}}_{3}q}{{\mathrm{log}}_{3}27}\\ =\frac{r}{{\mathrm{log}}_{3}{3}^{2}}-\frac{t}{{\mathrm{log}}_{3}{3}^{3}}\\ =\frac{r}{2{\mathrm{log}}_{3}3}-\frac{t}{3{\mathrm{log}}_{3}3}\\ =\frac{r}{2}-\frac{t}{3}\end{array}$
_{9}*p*– log_{27}*q***Soalan 4:**

(a)
Permudahkan:

log

_{2}(2*x*+ 1) – 5 log_{4}*x*^{2 }+ 4 log_{2}*x*
(b)
Seterusnya, selesaikan persamaan:

log

_{2}(2*x*+ 1) – 5 log_{4}*x*^{2 }+ 4 log_{2}*x*^{ }= 4

*Penyelesaian:***(a)**

log

_{2}(2*x*+ 1) – 5 log_{4}*x*^{2 }+ 4 log_{2}*x*$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5{\mathrm{log}}_{2}{x}^{2}}{{\mathrm{log}}_{2}4}+4{\mathrm{log}}_{2}x\\ ={\mathrm{log}}_{2}\left(2x+1\right)-\frac{5}{2}{\mathrm{log}}_{2}{x}^{2}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{\left({x}^{2}\right)}^{\left(\frac{5}{2}\right)}+{\mathrm{log}}_{2}{x}^{4}\end{array}$

$\begin{array}{l}={\mathrm{log}}_{2}\left(2x+1\right)-{\mathrm{log}}_{2}{x}^{5}+{\mathrm{log}}_{2}{x}^{4}\\ ={\mathrm{log}}_{2}\frac{\left(2x+1\right)\left({x}^{4}\right)}{{x}^{5}}\\ ={\mathrm{log}}_{2}\frac{2x+1}{x}\end{array}$

**(b)**

log

_{2}(2*x*+ 1) – 5 log_{4}*x*^{2 }+ 4 log_{2}*x*^{ }= 4$\begin{array}{l}{\mathrm{log}}_{2}\frac{2x+1}{x}=4\\ \text{}\frac{2x+1}{x}={2}^{4}\\ \text{}\frac{2x+1}{x}=16\\ \text{}2x+1=16x\\ \text{14}x=1\\ \text{}x=\frac{1}{14}\end{array}$