Sec 4 E Maths: Indices and Standard Form, practice questions & worked solutions

$$\begin{aligned}\dfrac4{128}&=\dfrac1{32}=2^{-5}\\\therefore a&=-5\end{aligned}$$

Answer: $a=-5$

$$\begin{aligned}&1\times10^{-3}\text{ s represent }70\times10^{-12}\text{ m}\\&1\text{ s represent }\dfrac{70\times10^{-12}}{10^{-3}}\text{ m}\\&1\text{ s represent }7\times10^{-8}\text{ m}\end{aligned}$$

Answer: $7\times10^{-8}$ m

(a)

$7.28\times10^{-3}$

(b)

(i) $81.7\times10^{500}$

(ii)

$$\begin{aligned}(6.3\times10^{500})+(8.17\times10^{501})&=(6.3+81.7)\times10^{500}\\&=88\times10^{500}\\&=8.8\times10^{501}\end{aligned}$$

Answer: $\text{(a) }7.28\times10^{-3}\quad\text{(b)(i) }81.7\times10^{500}\quad\text{(ii) }8.8\times10^{501}$

$$\begin{aligned}40\times7^4+9\times49^2&=7^k\\40\times7^4+9\times(7^2)^2&=7^k\\40\times7^4+9\times7^4&=7^k\\49\times7^4&=7^k\\7^2\times7^4&=7^k\\7^{2+4}&=7^k\\7^6&=7^k\\k&=6\end{aligned}$$

Answer: $k=6$

(a)

$1.90\times10^{27}\div5.97\times10^{24}=3.18\times10^2$

(b)

$$\begin{aligned}\text{Jupiter density}&=1.90\times10^{27}\div1.43\times10^{24}=1328.6713\text{ kg/m}^3\\\text{Earth density}&=5.97\times10^{24}\div1.08\times10^{21}=5527.7778\text{ kg/m}^3\end{aligned}$$

Earth has a higher density

Answer: (a) $3.18\times10^2$ (b) Earth has a higher density

(a)

$x^4$

(b)(i)

$$\begin{aligned}3^{-4}&=3^k\\k&=-4\end{aligned}$$

(b)(ii)

$$\begin{aligned}5^{4p-q}&=5^{3r}\\4p-q&=3r\\q&=4p-3r\end{aligned}$$

Answer: (a) $x^4$ (b)(i) $k=-4$ (b)(ii) $q=4p-3r$

(a)

$$\begin{aligned}2^{m-1}\times3^m&=1\\2^m\times3^m&=2\\6^m&=2\\36^m&=2^2\\36^m&=4\end{aligned}$$

(b)

$$\frac{(2xy)^{-2}}{12}\div\frac{x^3}{4y^2}=\frac1{48x^2y^2}\times\frac{4y^2}{x^3}=\frac1{12x^5}.$$

Answer: (a) $36^m=4$ (b) $\dfrac1{12x^5}$

(a)

$4.2\times10^6$

(b)

$$\begin{aligned}&\frac{4.2\times10^6-3.6\times10^6}{3.6\times10^6}\times100\%\\&=16\tfrac23\%\end{aligned}$$

(c)

$$\begin{aligned}&\frac{3.6\times10^6}{100-4}\times100\%\\&=3750000\\&=3.75\times10^6\end{aligned}$$

Answer: (a) $4.2\times10^6$ (b) $16\dfrac23\%$ (c) $3.75\times10^6$

(a)

$$\begin{aligned}&\frac{(2x^3y^2)^{-4}}{(10x^{-2}y^3)^2}\div\sqrt[3]{27x^{-3}y^6}\\[10pt]&=\frac{2^{-4}x^{-12}y^{-8}}{100x^{-4}y^6}\times\frac1{(27x^{-3}y^6)^{\frac13}}\\[10pt]&=\frac1{1600x^8y^{14}}\times\frac1{3x^{-1}y^2}\\[10pt]&=\frac1{4800x^7y^{16}}\end{aligned}$$

(b)

$$\begin{aligned}&3^{x+2}\times3(3^5)=1\\[0pt]&3^{x+2}\times3^1(3^5)=1\\[0pt]&3^{x+2}\times3^6=3^0\\[0pt]&x+2+6=0\\[0pt]&x=-8\end{aligned}$$

Answer: (a) $\dfrac1{4800x^7y^{16}}$ (b) $x=-8$

(a)

$$\begin{aligned}\frac{4^{\frac12}}{16^{y+1}}&=8^{2-y}\\\frac2{2^{4(y+1)}}&=2^{3(2-y)}\\\frac2{2^{4y+4}}&=2^{6-3y}\\2^{1-(4y+4)}&=2^{6-3y}\\1-(4y+4)&=6-3y\\1-4y-4&=6-3y\\1-4-6&=y\\y&=-9\end{aligned}$$

(b)

$$\begin{aligned}2^{500}&=\left(2^2\right)^{250}\\&=4^{250}\end{aligned}$$

$$\begin{aligned}4^{250}<5^{250}\end{aligned}$$

I disagree with her claim because $4^{250}<5^{250}$.

Answer: (a) $y=-9$ (b) disagree, as $2^{500}=4^{250}<5^{250}$