Question paper, Set 2

$3^{6}$ means $3$ multiplied by itself 6 times.

$3^{6} = 3 \times 3 \times 3 \times 3 \times 3 \times 3$

Answer: $3 \times 3 \times 3 \times 3 \times 3 \times 3$

$2^{7} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128$

$7^{2} = 7 \times 7 = 49$

$128 > 49$, so $2^{7}$ is greater.

Answer: $2^{7}$, since $128 > 49$

Divide repeatedly by $2$: $2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2$.

$2048 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

$= 2^{11}$

Answer: $2048 = 2^{11}$

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

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

$= 512$

Answer: $2^{9} = 512$

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

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

$= 7^{12}$

Answer: $7^{12}$

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

$2^{6} \times 5^{6} = (2 \times 5)^{6}$

$= (10)^{6} = 10 \times 10 \times 10 \times 10 \times 10 \times 10$

$= 10{,}00{,}000$

Answer: $(10)^{6} = 10{,}00{,}000$

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

$(x + 9)^{0} = 1$

Answer: $1$

Divide repeatedly by $2$: $32, 16, 8, 4, 2$, so $32 = 2^{5}$.

$\displaystyle \frac{1}{32} = \frac{1}{2^{5}}$

$= 2^{-5}$ (since $\displaystyle \frac{1}{a^n} = a^{-n}$)

Answer: $2^{-5}$

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

$\displaystyle \left(\frac{4}{7}\right)^{-6} = \left(\frac{7}{4}\right)^{6}$

Answer: $\displaystyle \left(\frac{7}{4}\right)^{6}$

$24$ is even, so $(-1)^{24} = 1$; $35$ is odd, so $(-1)^{35} = -1$.

$(-1)^{24} + (-1)^{35} = 1 + (-1) = 0$

Answer: $0$

Using $(-a)^n = -a^n \text{ if } n \text{ is odd}, \; (-a)^n = a^n \text{ if } n \text{ is even}$

$(-2)^{8} = 2^{8} = 256$ ($8$ is even)

$(-2)^{9} = -2^{9} = -512$ ($9$ is odd)

Answer: $(-2)^{8} = 256$, $(-2)^{9} = -512$

Given, $2^{2x-1} = 128$

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

$\Rightarrow 2^{2x-1} = 2^{7}$

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

$\Rightarrow 2x-1 = 7$

$\Rightarrow 2x = 7 + 1$

$\Rightarrow 2x = 8$

$\displaystyle \Rightarrow x = \frac{8}{2} = 4$

Answer: $x = 4$

Using $a^{mn} = (a^m)^n$

$5^{2x} = (5^{2})^{x}$

$= 25^{x}$

Answer: $25^{x}$

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

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

$\displaystyle = \frac{p^{9}}{27}$

Answer: $\displaystyle \frac{p^{9}}{27}$

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

$\displaystyle \frac{(3^{2})^{4} \times 3^{5}}{3^{9}} = \frac{3^{8} \times 3^{5}}{3^{9}} = \frac{3^{13}}{3^{9}}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$: $3^{13-9} = 3^{4}$

Answer: $3^{4}$

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

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

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

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

(i) Given, $5^{x+4} \div 5^{-2} = 625$

Dividing $625$ repeatedly by $5$: $625, 125, 25, 5$, so $625 = 5^{4}$.

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

$\Rightarrow 5^{x+6} = 5^{4}$

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

$\Rightarrow x+6 = 4$

$\Rightarrow x = 4 - 6$

$\Rightarrow x = -2$

(ii) Given, $(2^{3})^{x} \times 2 = 1024$

Dividing $1024$ repeatedly by $2$: $1024, 512, 256, 128, 64, 32, 16, 8, 4, 2$, so $1024 = 2^{10}$.

$\Rightarrow 2^{3x} \times 2^{1} = 2^{10}$ ($(a^m)^n = a^{m \times n}$)

$\Rightarrow 2^{3x+1} = 2^{10}$ ($a^m \times a^n = a^{m+n}$)

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

$\Rightarrow 3x+1 = 10$

$\Rightarrow 3x = 10 - 1$

$\Rightarrow 3x = 9$

$\displaystyle \Rightarrow x = \frac{9}{3} = 3$

Answer: (i) $x = -2$ (ii) $x = 3$

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

$\displaystyle 2^{5} \times \left(\frac{3}{2}\right)^{5} = \left(2 \times \frac{3}{2}\right)^{5}$

$= 3^{5} = 243$

(ii) Using $(-1)^{\text{odd number}} = -1, \; (-1)^{\text{even number}} = 1$

$(-1)^{4} + (-1)^{7} + (-1)^{9} = 1 + (-1) + (-1)$

$= -1$

Answer: (i) $243$ (ii) $-1$

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

$\displaystyle 2^{\frac{11}{2}} \div 2^{\frac{5}{2}} = 2^{\frac{11}{2} - \frac{5}{2}}$

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

$= 8$

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

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

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

$= 16$

Answer: (i) $8$ (ii) $16$

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

$\displaystyle 9^{2} \times \left(\frac{5}{3}\right)^{4} = 81 \times \frac{5^{4}}{3^{4}}$

$\displaystyle = 81 \times \frac{625}{81}$

$= 625$

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

$\displaystyle = \frac{3 - 2}{6} \div \frac{1}{5} = \frac{1}{6} \div \frac{1}{5}$

$\displaystyle = \frac{1}{6} \times 5 = \frac{5}{6}$

Answer: (i) $625$ (ii) $\displaystyle \frac{5}{6}$