Question paper, Set 5

$\displaystyle \frac{2}{3}$ is multiplied by itself 4 times.

$\displaystyle \therefore \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} \times \frac{2}{3} = \left(\frac{2}{3}\right)^{4}$, read as '$\displaystyle \frac{2}{3}$ power 4'

Answer: $\displaystyle \left(\frac{2}{3}\right)^{4}$

$15^{3} = 15 \times 15 \times 15$

$= 3375$

Answer: $3375$

Divide repeatedly by $6$: $7776, 1296, 216, 36, 6$.

$7776 = 6 \times 6 \times 6 \times 6 \times 6$

$= 6^{5}$

Answer: $7776 = 6^{5}$

Using $a^m \times a^n = a^{m+n}$

$11^{5} \times 11^{8} = 11^{5+8}$

$= 11^{13}$

Answer: $11^{13}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{5^{9}}{5^{6}} = 5^{9-6} = 5^{3}$

$= 125$

Answer: $5^{3} = 125$

Using $a^n \times b^n = (a \times b)^n$

$2^{3} \times 9^{3} = (2 \times 9)^{3}$

$= (18)^{3} = 18 \times 18 \times 18$

$= 5832$

Answer: $(18)^{3} = 5832$

Using $\displaystyle \frac{a^n}{b^n} = \left(\frac{a}{b}\right)^n$

$\displaystyle \frac{18^{4}}{6^{4}} = \left(\frac{18}{6}\right)^{4}$

$= 3^{4}$

$= 81$

Answer: $81$

$8^{0} = 1$ and $13^{0} = 1$

$8^{0} \times 13^{0} = 1 \times 1 = 1$

Answer: $1$

Using $\displaystyle a^{-n} = \frac{1}{a^n}$

$\displaystyle 10^{-3} = \frac{1}{10^{3}}$

$\displaystyle = \frac{1}{1000}$

Answer: $\displaystyle \frac{1}{1000}$

$131$ is an odd number.

$(-1)^{\text{odd number}} = -1$

$\therefore (-1)^{131} = -1$

Answer: $-1$

$4$ is even, and $(-a)^n = a^n$ for even $n$.

$(-4)^{4} = 4^{4} = 256$

Answer: $256$

Given, $4^{x-2} = 256$

Dividing $256$ repeatedly by $4$: $256, 64, 16, 4$, so $256 = 4^{4}$.

$\Rightarrow 4^{x-2} = 4^{4}$

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x-2 = 4$

$\Rightarrow x = 4 + 2$

$\Rightarrow x = 6$

Answer: $x = 6$

Using $\displaystyle \left(\frac{b}{a}\right)^n = \left(\frac{a}{b}\right)^{-n}$: $\displaystyle \left(\frac{3}{5}\right)^{4} = \left(\frac{5}{3}\right)^{-4}$

Using $\displaystyle \frac{1}{a^n} = a^{-n}$: $\displaystyle \frac{1}{11^{3}} = 11^{-3}$

Answer: $\displaystyle \left(\frac{5}{3}\right)^{-4}$, $11^{-3}$

Using $\displaystyle \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$ and $(a^m)^n = a^{m \times n}$

$\displaystyle \left(\frac{4}{z^{2}}\right)^{3} = \frac{4^{3}}{(z^{2})^{3}}$

$\displaystyle = \frac{64}{z^{6}}$

Answer: $\displaystyle \frac{64}{z^{6}}$

Using $(a^m)^n = a^{m \times n}$ and $a^m \times a^n = a^{m+n}$

$\displaystyle \frac{(5^{3})^{2} \times 5^{4}}{5^{7}} = \frac{5^{6} \times 5^{4}}{5^{7}} = \frac{5^{10}}{5^{7}}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$: $5^{10-7} = 5^{3}$

Answer: $5^{3}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle 2^{3} \div 2^{-4} = \frac{2^{3}}{2^{-4}} = 2^{3-(-4)} = 2^{3+4} = 2^{7}$

$= 128$

Answer: $128$

(i) Given, $(5^{2})^{x} \times 5^{-1} = 125$

Dividing $125$ repeatedly by $5$: $125, 25, 5$, so $125 = 5^{3}$.

$\Rightarrow 5^{2x} \times 5^{-1} = 5^{3}$ ($(a^m)^n = a^{m \times n}$)

$\Rightarrow 5^{2x-1} = 5^{3}$ ($a^m \times a^n = a^{m+n}$)

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow 2x-1 = 3$

$\Rightarrow 2x = 3 + 1$

$\Rightarrow 2x = 4$

$\displaystyle \Rightarrow x = \frac{4}{2} = 2$

(ii) Given, $(-2)^{x-1} \times (-2)^{3} = (-2)^{7}$

$\Rightarrow (-2)^{x-1+3} = (-2)^{7}$ ($a^m \times a^n = a^{m+n}$)

Using $a^m = a^n \Rightarrow m = n$

$\Rightarrow x+2 = 7$

$\Rightarrow x = 7 - 2$

$\Rightarrow x = 5$

Answer: (i) $x = 2$ (ii) $x = 5$

(i) Using $a^n \times b^n = (a \times b)^n$

$\displaystyle 5^{6} \times \left(\frac{2}{5}\right)^{6} = \left(5 \times \frac{2}{5}\right)^{6}$

$= 2^{6} = 64$

(ii) Using $\displaystyle \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n$

$\displaystyle \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{6}\right)^{-2} = (2)^{2} + (5)^{2} + (6)^{2}$

$= 4 + 25 + 36$

$= 65$

Answer: (i) $64$ (ii) $65$

(i) Using $a^m \times a^n = a^{m+n}$

$\displaystyle 7^{\frac{1}{2}} \times 7^{\frac{3}{2}} = 7^{\frac{1}{2} + \frac{3}{2}}$

$\displaystyle = 7^{\frac{4}{2}} = 7^{2}$

$= 49$

(ii) Using $a^m \times a^n = a^{m+n}$ and $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{7^{\frac{7}{2}} \times 7^{-\frac{1}{2}}}{7^{4}} = \frac{7^{\frac{7}{2} + (-\frac{1}{2})}}{7^{4}}$

$\displaystyle = \frac{7^{\frac{6}{2}}}{7^{4}} = \frac{7^{3}}{7^{4}}$

$= 7^{3-4} = 7^{-1}$

$\displaystyle = \frac{1}{7^{1}} = \frac{1}{7}$

Answer: (i) $49$ (ii) $\displaystyle \frac{1}{7}$

(i) $\displaystyle \left(2^{7} \div 2^{4}\right)^{2} \times 2^{-3} = \left(\frac{2^{7}}{2^{4}}\right)^{2} \times 2^{-3}$

$= (2^{7-4})^{2} \times 2^{-3} = (2^{3})^{2} \times 2^{-3}$

$= 2^{6} \times 2^{-3}$ ($(a^m)^n = a^{m \times n}$)

$= 2^{6-3} = 2^{3}$

$= 8$

(ii) $\displaystyle \left(3^{-1} - 4^{-1}\right) \div 2^{-1} = \left(\frac{1}{3} - \frac{1}{4}\right) \div \frac{1}{2}$

$\displaystyle = \frac{4 - 3}{12} \div \frac{1}{2} = \frac{1}{12} \div \frac{1}{2}$

$\displaystyle = \frac{1}{12} \times 2 = \frac{1}{6}$

Answer: (i) $8$ (ii) $\displaystyle \frac{1}{6}$