Soalan 5:
Diberi bahawa logmp = x dan lognp = y.
Menggunakan hukum-hukum logaritma, ungkapkan yang berikut dalam sebutan x dan y.
$$ \log _{\frac{m}{n}} p $$
Jawapan:
$$ \begin{aligned} \log _{\frac{m}{n}} p & =\frac{1}{\log _p \frac{m}{n}} \\ & =\frac{1}{\log _p m-\log _p n} \\ & =\frac{1}{\frac{1}{\log _m p}-\frac{1}{\log _n p}} \\ & =\frac{1}{\frac{1}{x}-\frac{1}{y}} \\ & =\frac{1}{\frac{y-x}{x y}} \\ & =\frac{x y}{y-x} \end{aligned} $$
Diberi bahawa logmp = x dan lognp = y.
Menggunakan hukum-hukum logaritma, ungkapkan yang berikut dalam sebutan x dan y.
$$ \log _{\frac{m}{n}} p $$
Jawapan:
$$ \begin{aligned} \log _{\frac{m}{n}} p & =\frac{1}{\log _p \frac{m}{n}} \\ & =\frac{1}{\log _p m-\log _p n} \\ & =\frac{1}{\frac{1}{\log _m p}-\frac{1}{\log _n p}} \\ & =\frac{1}{\frac{1}{x}-\frac{1}{y}} \\ & =\frac{1}{\frac{y-x}{x y}} \\ & =\frac{x y}{y-x} \end{aligned} $$
Soalan 6:
Selesaikan persamaan serentak berikut dengan menggunakan kaedah penggantian dan / atau penghapusan.
$$ 2 x+3 y-2 z=-6 \quad, \quad-3 x+y+4 z=-7 \quad, \quad-4 x-5 y-6 z=4 $$
[5 markah]
Jawapan:
$$ \begin{aligned} (2 x+3 y-2 z & =-6) \times 6 \\ 12 x+18 y-12 z & =-36 \ldots(1) \end{aligned} $$
$$ \begin{aligned} (-3 x+y+4 z & =-7) \times 3 \\ -9 x+3 y+12 z & =-21 \ldots(2) \end{aligned} $$
$$ \begin{array}{r} (-4 x-5 y-6 z=4) \times 2 \\ -8 x-10 y-12 z=8 \ldots (3) \end{array} $$
$$ \begin{aligned} &\text { (1) }+ \text { (2): }\\ &\begin{aligned} (12 x+18 y-12 z)+(-9 x+3 y+12 z) & =-36+(-21) \\ 3 x+21 y & =-57 \\ x+7 y & =-19 \\ x & =-19-7 y \ldots (4) \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { (1) }- \text { (3): }\\ &\begin{aligned} (12 x+18 y-12 z)-(-8 x-10 y-12 z) & =-36-8 \\ 20 x+28 y & =-44 \\ 5 x+7 y & =-11 \\ 5(-19-7 y)+7 y & =-11 \\ -28 y & =-11+95 \\ -28 y & =84 \\ y & =\frac{84}{-28} \\ y & =-3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Dari (4), }\\ &\begin{aligned} & x=-19-7(-3) \\ & x=-19+21 \\ & x=2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Dari (1), }\\ &\begin{aligned} 12(2)+18(-3)-12 z & =-36 \\ 24-54-12 z & =-36 \\ -12 z & =-36-24+54 \\ -12 z & =-6 \\ z & =\frac{-6}{-12} \\ z & =\frac{1}{2} \end{aligned} \end{aligned} $$
$$ \therefore x=2, y=-3, z=\frac{1}{2} $$
Selesaikan persamaan serentak berikut dengan menggunakan kaedah penggantian dan / atau penghapusan.
$$ 2 x+3 y-2 z=-6 \quad, \quad-3 x+y+4 z=-7 \quad, \quad-4 x-5 y-6 z=4 $$
[5 markah]
Jawapan:
$$ \begin{aligned} (2 x+3 y-2 z & =-6) \times 6 \\ 12 x+18 y-12 z & =-36 \ldots(1) \end{aligned} $$
$$ \begin{aligned} (-3 x+y+4 z & =-7) \times 3 \\ -9 x+3 y+12 z & =-21 \ldots(2) \end{aligned} $$
$$ \begin{array}{r} (-4 x-5 y-6 z=4) \times 2 \\ -8 x-10 y-12 z=8 \ldots (3) \end{array} $$
$$ \begin{aligned} &\text { (1) }+ \text { (2): }\\ &\begin{aligned} (12 x+18 y-12 z)+(-9 x+3 y+12 z) & =-36+(-21) \\ 3 x+21 y & =-57 \\ x+7 y & =-19 \\ x & =-19-7 y \ldots (4) \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { (1) }- \text { (3): }\\ &\begin{aligned} (12 x+18 y-12 z)-(-8 x-10 y-12 z) & =-36-8 \\ 20 x+28 y & =-44 \\ 5 x+7 y & =-11 \\ 5(-19-7 y)+7 y & =-11 \\ -28 y & =-11+95 \\ -28 y & =84 \\ y & =\frac{84}{-28} \\ y & =-3 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Dari (4), }\\ &\begin{aligned} & x=-19-7(-3) \\ & x=-19+21 \\ & x=2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} &\text { Dari (1), }\\ &\begin{aligned} 12(2)+18(-3)-12 z & =-36 \\ 24-54-12 z & =-36 \\ -12 z & =-36-24+54 \\ -12 z & =-6 \\ z & =\frac{-6}{-12} \\ z & =\frac{1}{2} \end{aligned} \end{aligned} $$
$$ \therefore x=2, y=-3, z=\frac{1}{2} $$