Cabaran Dinamis 1 (Uji Diri) – Buku Teks Matematik Tingkatan 3 Bab 1 (Indeks)

Soalan 1:
Nyatakan sama ada operasi yang melibatkan hukum indeks berikut benar atau palsu. Jika palsu, nyatakan jawapan yang betul.


Penyelesaian:
(a)
$$\begin{aligned} &a^5=a \times a \times a \times a \times a\\ &\text { berulang lima kali } \end{aligned}$$
$$\begin{aligned} &a^5=a^5\\ &\text { Benar } \end{aligned}$$
(b)
$$\begin{aligned} & 5^2=10 \\ & 5^2=5 \times 5 \\ &=25 \\ & \neq 10 \\ & \text { Palsu } \end{aligned}$$

(c)
$$\begin{aligned} 3^0 & =0 \quad \text {,daripada formula, } a^0=1 \\ \therefore 3^0 & =1 \\ & \neq 0 \\ & \text { Palsu } \end{aligned}$$
(d)
$$\begin{aligned} \left(2 x^3\right)^5 & =2 x^{15} \\ \left(2 x^3\right)^5 & =2^5 x^{15} \\ & =32 x^{15} \\ & \neq 2 x^{15} \\ & \text { Palsu } \end{aligned}$$
(e)
$$\begin{aligned} &m^0 n^0=1\\ &\text { Benar } \end{aligned}$$
(f)
$$\begin{aligned} & 2 a^{-4}=\frac{1}{2 a^4} \\ & 2 a^{-4}=\frac{2}{a^4} \\ & \neq \frac{1}{2 a^4} \\ & \text { Palsu } \end{aligned}$$

(g)
$$\begin{aligned} &\begin{aligned} 32^{\frac{2}{5}} & =(\sqrt[2]{32})^5 \\ 32^{\frac{2}{5}} & =\left(32^{\frac{1}{5}}\right)^2 \\ & =(\sqrt[5]{32})^2 \\ & \neq(\sqrt[2]{32})^5 \end{aligned}\\ &\text { Palsu } \end{aligned}$$
(h)
$$\begin{aligned} &\left(\frac{m}{n}\right)^{-4}=\left(\frac{n}{m}\right)^4\\ &\text { Benar } \end{aligned}$$
(i)
$$\begin{aligned} \left(5 m^{\frac{1}{4}}\right)^{-4} & =\frac{625}{m} \\ \left(5 m^{\frac{1}{4}}\right)^{-4} & =5^{-4} m^{\frac{1}{4} \times(-4)} \\ & =\frac{1}{5^4} \times m^{-1} \\ & =\frac{1}{625} \times \frac{1}{m} \\ & =\frac{1}{625 m} \\ & \neq \frac{625}{m} \\ & P a l s u \end{aligned}$$

Soalan 2:
Salin dan lengkapkan rajah di bawah dengan nilai yang sesuai.


Penyelesaian:
$$\begin{aligned} 5^{(x)} \times 5^5 & =5^9 \\ 5^{x+5} & =5^9 \\ x+5 & =9 \\ x & =9-5 \\ x & =4 \end{aligned}$$
$$\begin{aligned} 5^{3(x)} & =5^9 \\ 3 x & =9 \\ x & =\frac{9}{3} \\ x & =3 \end{aligned}$$

$$\begin{aligned} 5^{12} \div 5^{(x)} & =5^9 \\ 5^{12-x} & =5^9 \\ 12-x & =9 \\ -x & =9-12 \\ -x & =-3 \\ x & =3 \end{aligned}$$
$$\begin{aligned} (\sqrt{25})^{(x)} & =5^9 \\ \left(25^{\frac{1}{2}}\right)^x & =5^9 \\ \left(5^{2 \times \frac{1}{2}}\right)^x & =5^9 \\ 5^x & =5^9 \\ x & =9 \end{aligned}$$
$$\begin{aligned} \left(\frac{1}{5}\right)^{(x)} & =5^9 \\ \left(5^{-1}\right)^{(x)} & =5^9 \\ 5^{-x} & =5^9 \\ -x & =9 \\ x & =-9 \end{aligned}$$

$$\begin{aligned} (x \sqrt{125})^y & =5^9 \\ \left(x \sqrt{5^3}\right)^y & =5^9 \\ \left(\left(5^3\right)^{\frac{1}{x}}\right)^y & =5^9 \\ 5^{\frac{3 y}{x}} & =5^9 \\ \frac{3 y}{x} & =9 \\ \frac{y}{x} & =\frac{9}{3} \\ y=9, x & =3 \end{aligned}$$
$$\begin{aligned} \left(5^{(x)}\right)^{\frac{3}{2}} & =5^9 \\ 5^{\frac{3}{2}} x & =5^9 \\ \frac{3}{2} x & =9 \\ x & =9 \times \frac{2}{3} \\ x & =6 \end{aligned}$$

$$\begin{aligned} \frac{5^6 \times 5^{(x)}}{5^2} & =5^9 \\ 5^{6+x-2} & =5^9 \\ 5^{4+x} & =5^9 \\ 4+x & =9 \\ x & =9-4 \\ x & =5 \end{aligned}$$
$$\begin{aligned} \frac{1}{5^{(x)}} & =5^9 \\ 5^{-x} & =5^9 \\ -x & =9 \\ x & =-9 \end{aligned}$$
$$\begin{aligned} \left(\frac{1}{5^{(x)}}\right)^3 & =5^9 \\ \left(5^{-x}\right)^3 & =5^9 \\ 5^{-3 x} & =5^9 \\ -3 x & =9 \\ x & =\frac{9}{-3} \\ x & =-3 \end{aligned}$$

Soalan 3:
Salin dan lengkapkan rajah di bawah.


Penyelesaian:


$$\begin{aligned} \left(5^{-1} \times \sqrt{25}\right)^3 & =\left[5^{-1} \times\left(5^2\right)^{\frac{1}{2}}\right]^3 \\ & =\left[5^{-1} \times 5\right]^3 \\ & =\left(5^{-1+1}\right)^3 \\ & =\left(5^0\right)^3 \\ & =5^0 \\ & =1 \end{aligned}$$