Question paper, Set 9

$y$ is multiplied by itself 4 times.

$\therefore y \times y \times y \times y = y^{4}$, read as '$y$ power 4'

Answer: $y^{4}$

$8^{4} = 8 \times 8 \times 8 \times 8$

$= 4096$

Answer: $4096$

Divide repeatedly by $5$: $15{,}625, 3125, 625, 125, 25, 5$.

$15{,}625 = 5 \times 5 \times 5 \times 5 \times 5 \times 5$

$= 5^{6}$

Answer: $15{,}625 = 5^{6}$

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

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

$\displaystyle = \left(\frac{3}{4}\right)^{9}$

Answer: $\displaystyle \left(\frac{3}{4}\right)^{9}$

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

$\displaystyle \frac{10^{9}}{10^{4}} = 10^{9-4} = 10^{5}$

$= 1{,}00{,}000$

Answer: $10^{5} = 1{,}00{,}000$

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

$25^{2} \times 4^{2} = (25 \times 4)^{2}$

$= (100)^{2} = 100 \times 100$

$= 10{,}000$

Answer: $(100)^{2} = 10{,}000$

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

$\displaystyle \frac{45^{5}}{15^{5}} = \left(\frac{45}{15}\right)^{5}$

$= 3^{5}$

$= 243$

Answer: $243$

$48 + 52 = 100 \ne 0$

$(48 + 52)^{0} = 100^{0} = 1$

Answer: $1$

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

$\displaystyle 7^{-2} = \frac{1}{7^{2}}$

$\displaystyle = \frac{1}{49}$

Answer: $\displaystyle \frac{1}{49}$

$999$ is an odd number.

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

$\therefore (-1)^{999} = -1$

Answer: $-1$

$3$ is odd, and $(-a)^n = -a^n$ for odd $n$.

$(-6)^{3} = -6^{3} = -216$

Answer: $-216$

Given, $3^{2x+5} = 27$

Dividing $27$ repeatedly by $3$: $27, 9, 3$, so $27 = 3^{3}$.

$\Rightarrow 3^{2x+5} = 3^{3}$

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

$\Rightarrow 2x+5 = 3$

$\Rightarrow 2x = 3 - 5$

$\Rightarrow 2x = -2$

$\displaystyle \Rightarrow x = \frac{-2}{2} = -1$

Answer: $x = -1$

Using $-a^n = (-a)^n$ for odd $n$: $-6^{5} = (-6)^{5}$

Using $a^n = (-a)^n$ for even $n$: $6^{4} = (-6)^{4}$

Answer: $-6^{5} = (-6)^{5}$, $6^{4} = (-6)^{4}$

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

$(3ab^{4})^{2} = 3^{2} \times a^{2} \times (b^{4})^{2}$

$= 9a^{2}b^{8}$

Answer: $9a^{2}b^{8}$

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

$\displaystyle \frac{10^{12} \times 10^{3}}{10^{9}} = \frac{10^{15}}{10^{9}}$

$= 10^{15-9} = 10^{6}$

Answer: $10^{6}$

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

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

$\displaystyle = \frac{1}{2^{6}} = \frac{1}{64}$

Answer: $\displaystyle \frac{1}{64}$

(i) Given, $3^{x-4} \times 3^{2} = 243$

Dividing $243$ repeatedly by $3$: $243, 81, 27, 9, 3$, so $243 = 3^{5}$.

$\Rightarrow 3^{x-4+2} = 3^{5}$ ($a^m \times a^n = a^{m+n}$)

$\Rightarrow 3^{x-2} = 3^{5}$

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

$\Rightarrow x-2 = 5$

$\Rightarrow x = 5 + 2$

$\Rightarrow x = 7$

(ii) Given, $7^{3x+1} \div 7^{-1} = 49$

Dividing $49$ repeatedly by $7$: $49, 7$, so $49 = 7^{2}$.

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

$\Rightarrow 7^{3x+2} = 7^{2}$

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

$\Rightarrow 3x+2 = 2$

$\Rightarrow 3x = 2 - 2$

$\Rightarrow 3x = 0$

$\displaystyle \Rightarrow x = \frac{0}{3} = 0$

Answer: (i) $x = 7$ (ii) $x = 0$

(i) $\displaystyle \left(9^{0} + 5^{-1}\right) \times 5^{2} = \left(1 + \frac{1}{5}\right) \times 25$

$\displaystyle = \frac{6}{5} \times 25$

$= 30$

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

$\displaystyle \left(\frac{1}{5}\right)^{-2} + \left(\frac{1}{6}\right)^{-2} + \left(\frac{1}{7}\right)^{-2} = (5)^{2} + (6)^{2} + (7)^{2}$

$= 25 + 36 + 49$

$= 110$

Answer: (i) $30$ (ii) $110$

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

$\displaystyle \frac{10^{\frac{9}{4}} \times 10^{-\frac{1}{4}}}{10^{3}} = \frac{10^{\frac{9}{4} + (-\frac{1}{4})}}{10^{3}}$

$\displaystyle = \frac{10^{\frac{8}{4}}}{10^{3}} = \frac{10^{2}}{10^{3}}$

$= 10^{2-3} = 10^{-1}$

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

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

$\displaystyle 10^{1} \times 10^{\frac{4}{3}} \times 10^{-\frac{1}{3}} = 10^{1 + \frac{4}{3} + (-\frac{1}{3})}$

$\displaystyle = 10^{\frac{6}{3}} = 10^{2}$

$= 100$

Answer: (i) $\displaystyle \frac{1}{10}$ (ii) $100$

(i) Using $(ab)^n = a^n \times b^n$ and $(a^m)^n = a^{m \times n}$

$(2m^{4})^{3} \times (3m)^{2} = 2^{3} \cdot (m^{4})^{3} \times 3^{2} \cdot m^{2}$

$= 8m^{12} \times 9m^{2}$

$= 72 \cdot m^{12+2} = 72m^{14}$

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

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

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

$= 3^{2} \times 2^{-1} \times 7^{-1}$ ($a^m \times a^n = a^{m+n}$, $\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

$\displaystyle = 9 \times \frac{1}{2} \times \frac{1}{7}$

$\displaystyle = \frac{9}{14}$

Answer: (i) $72m^{14}$ (ii) $\displaystyle \frac{9}{14}$