Question paper, Set 1

$7$ is multiplied by itself 5 times.

$\therefore 7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$, read as '$7$ power 5'

Answer: $7^{5}$

$12^{3} = 12 \times 12 \times 12$

$= 1728$

Answer: $1728$

Divide repeatedly by $3$: $729, 243, 81, 27, 9, 3$.

$729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$= 3^{6}$

Answer: $729 = 3^{6}$

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

$6^{4} \times 6^{9} = 6^{4+9}$

$= 6^{13}$

Answer: $6^{13}$

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

$\displaystyle \frac{2^{12}}{2^{5}} = 2^{12-5} = 2^{7}$

$= 128$

Answer: $2^{7} = 128$

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

$4^{3} \times 5^{3} = (4 \times 5)^{3}$

$= (20)^{3} = 20 \times 20 \times 20$

$= 8000$

Answer: $(20)^{3} = 8000$

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

$\displaystyle \frac{12^{3}}{4^{3}} = \left(\frac{12}{4}\right)^{3}$

$= 3^{3}$

$= 27$

Answer: $27$

Using $a^0 = 1 \; (a \ne 0)$

$27^{0} = 1$

Answer: $1$

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

$\displaystyle 3^{-4} = \frac{1}{3^{4}}$

$\displaystyle = \frac{1}{81}$

Answer: $\displaystyle \frac{1}{81}$

$57$ is an odd number.

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

$\therefore (-1)^{57} = -1$

Answer: $-1$

$7$ is odd, and $(-a)^n = -a^n$ for odd $n$.

$(-2)^{7} = -2^{7} = -128$

Answer: $-128$

Given, $3^{x+4} = 243$

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

$\Rightarrow 3^{x+4} = 3^{5}$

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

$\Rightarrow x+4 = 5$

$\Rightarrow x = 5 - 4$

$\Rightarrow x = 1$

Answer: $x = 1$

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

$4^{x+3} = 4^{x} \times 4^{3}$

$= 64 \times 4^{x}$

Answer: $64 \times 4^{x}$

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

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

$= 16 \times a^{12} = 16a^{12}$

Answer: $16a^{12}$

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

$\displaystyle \frac{3^{4} \times 3^{9}}{3^{7}} = \frac{3^{13}}{3^{7}}$

$= 3^{13-7} = 3^{6}$

Answer: $3^{6}$

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

$2^{-5} \times 2^{9} = 2^{-5+9} = 2^{4}$

$= 16$

Answer: $16$

(i) Given, $2^{x+1} \times 2^{4} = 128$

Dividing $128$ repeatedly by $2$: $128, 64, 32, 16, 8, 4, 2$, so $128 = 2^{7}$.

$\Rightarrow 2^{x+1+4} = 2^{7}$ ($a^m \times a^n = a^{m+n}$)

$\Rightarrow 2^{x+5} = 2^{7}$

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

$\Rightarrow x+5 = 7$

$\Rightarrow x = 7 - 5$

$\Rightarrow x = 2$

(ii) Given, $\displaystyle 3^{2m+1} = \frac{1}{27}$

$\displaystyle \frac{1}{27} = \frac{1}{3^{3}} = 3^{-3}$ (dividing $27$ repeatedly by $3$: $27, 9, 3$)

$\Rightarrow 3^{2m+1} = 3^{-3}$

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

$\Rightarrow 2m+1 = -3$

$\Rightarrow 2m = -3 - 1$

$\Rightarrow 2m = -4$

$\displaystyle \Rightarrow m = \frac{-4}{2} = -2$

Answer: (i) $x = 2$ (ii) $m = -2$

(i) $\displaystyle \left(3^{12} \div 3^{9}\right) \times 3^{-1} = \frac{3^{12}}{3^{9}} \times 3^{-1}$

$= 3^{12-9} \times 3^{-1} = 3^{3} \times 3^{-1}$

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

$= 9$

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

$\displaystyle \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} + \left(\frac{1}{5}\right)^{-2} = (3)^{2} + (4)^{2} + (5)^{2}$

$= 9 + 16 + 25$

$= 50$

Answer: (i) $9$ (ii) $50$

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

$\displaystyle 3^{\frac{5}{2}} \times 3^{\frac{3}{2}} = 3^{\frac{5}{2} + \frac{3}{2}}$

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

$= 81$

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

$\displaystyle \frac{3^{\frac{9}{2}} \times 3^{-\frac{3}{2}}}{3^{2}} = \frac{3^{\frac{9}{2} + (-\frac{3}{2})}}{3^{2}}$

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

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

$= 3$

Answer: (i) $81$ (ii) $3$

(i) $\displaystyle \left(3^{4} \div 3^{6}\right)^{3} \times 3^{2} = \left(\frac{3^{4}}{3^{6}}\right)^{3} \times 3^{2}$

$= (3^{4-6})^{3} \times 3^{2} = (3^{-2})^{3} \times 3^{2}$

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

$= 3^{-6+2} = 3^{-4}$

$\displaystyle = \frac{1}{3^{4}} = \frac{1}{81}$

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

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

$= 27x^{6} \times 4x^{2}$

$= 108 \cdot x^{6+2} = 108x^{8}$

Answer: (i) $\displaystyle \frac{1}{81}$ (ii) $108x^{8}$