Question paper, Set 6

$8^{5}$ means $8$ multiplied by itself 5 times.

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

Answer: $8 \times 8 \times 8 \times 8 \times 8$

$2^{9} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 512$

$9^{2} = 9 \times 9 = 81$

$512 > 81$, so $2^{9}$ is greater.

Answer: $2^{9}$, since $512 > 81$

Divide repeatedly by $3$: $6561, 2187, 729, 243, 81, 27, 9, 3$.

$6561 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$= 3^{8}$

Answer: $6561 = 3^{8}$

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

$10^{2} \times 10^{4} = 10^{2+4} = 10^{6}$

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

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

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

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

$\displaystyle = \left(\frac{2}{9}\right)^{12}$

Answer: $\displaystyle \left(\frac{2}{9}\right)^{12}$

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

$4^{2} \times 9^{2} = (4 \times 9)^{2}$

$= (36)^{2} = 36 \times 36$

$= 1296$

Answer: $(36)^{2} = 1296$

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

$-14 \ne 0$, so $(-14)^{0} = 1$

Answer: $1$

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

$\displaystyle \frac{1}{256} = \frac{1}{4^{4}}$

$= 4^{-4}$ (since $\displaystyle \frac{1}{a^n} = a^{-n}$)

Answer: $4^{-4}$

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

$\displaystyle \left(\frac{11}{3}\right)^{-4} = \left(\frac{3}{11}\right)^{4}$

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

$63$ is odd, so $(-1)^{63} = -1$; $71$ is odd, so $(-1)^{71} = -1$.

$(-1)^{63} + (-1)^{71} = -1 + (-1) = -2$

Answer: $-2$

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

$(-3)^{6} = 3^{6} = 729$ ($6$ is even)

$(-3)^{7} = -3^{7} = -2187$ ($7$ is odd)

Answer: $(-3)^{6} = 729$, $(-3)^{7} = -2187$

Given, $6^{3n-2} = 216$

Dividing $216$ repeatedly by $6$: $216, 36, 6$, so $216 = 6^{3}$.

$\Rightarrow 6^{3n-2} = 6^{3}$

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

$\Rightarrow 3n-2 = 3$

$\Rightarrow 3n = 3 + 2$

$\Rightarrow 3n = 5$

$\displaystyle \Rightarrow n = \frac{5}{3}$

Answer: $\displaystyle n = \frac{5}{3}$

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

$10^{x+3} = 10^{x} \times 10^{3}$

$= 1000 \times 10^{x}$

Answer: $1000 \times 10^{x}$

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

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

$= 36 \times x^{4} = 36x^{4}$

Answer: $36x^{4}$

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

$(x^{2}y)^{2} \times (xy^{3})^{3} = x^{4}y^{2} \times x^{3}y^{9}$

Using $a^m \times a^n = a^{m+n}$: $= x^{7}y^{11}$

Answer: $x^{7}y^{11}$

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

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

$= 4096$

Answer: $4096$

(i) Given, $10^{2x+1} \times 10^{1} = 1{,}00{,}000$

Dividing $1{,}00{,}000$ repeatedly by $10$: $1{,}00{,}000, 10{,}000, 1000, 100, 10$, so $1{,}00{,}000 = 10^{5}$.

$\Rightarrow 10^{2x+1+1} = 10^{5}$ ($a^m \times a^n = a^{m+n}$)

$\Rightarrow 10^{2x+2} = 10^{5}$

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

$\Rightarrow 2x+2 = 5$

$\Rightarrow 2x = 5 - 2$

$\Rightarrow 2x = 3$

$\displaystyle \Rightarrow x = \frac{3}{2}$

(ii) Given, $\displaystyle 5^{x-2} = \frac{1}{25}$

$\displaystyle \frac{1}{25} = \frac{1}{5^{2}} = 5^{-2}$ (dividing $25$ repeatedly by $5$: $25, 5$)

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

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

$\Rightarrow x-2 = -2$

$\Rightarrow x = -2 + 2$

$\Rightarrow x = 0$

Answer: (i) $\displaystyle x = \frac{3}{2}$ (ii) $x = 0$

(i) $\displaystyle \left(7^{0} + 2^{-1}\right) \times 4^{2} = \left(1 + \frac{1}{2}\right) \times 16$

$\displaystyle = \frac{3}{2} \times 16$

$= 24$

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

$(-1)^{6} + (-1)^{10} + (-1)^{13} = 1 + 1 + (-1)$

$= 1$

Answer: (i) $24$ (ii) $1$

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

$\displaystyle 5^{\frac{2}{3}} \times 5^{\frac{4}{3}} = 5^{\frac{2}{3} + \frac{4}{3}}$

$\displaystyle = 5^{\frac{6}{3}} = 5^{2}$

$= 25$

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

$\displaystyle 3^{\frac{1}{2}} \div 3^{\frac{9}{2}} = 3^{\frac{1}{2} - \frac{9}{2}}$

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

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

Answer: (i) $25$ (ii) $\displaystyle \frac{1}{81}$

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

$(5a)^{2} \times (2a)^{3} = 5^{2} \cdot a^{2} \times 2^{3} \cdot a^{3}$

$= 25a^{2} \times 8a^{3}$

$= 200 \cdot a^{2+3} = 200a^{5}$

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

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

$\displaystyle = \frac{3^{3}}{2^{3}} \times \frac{2^{2}}{5^{2}} \times \frac{5^{1}}{3^{1}}$

$= 2^{-1} \times 3^{2} \times 5^{-1}$ ($a^m \times a^n = a^{m+n}$, $\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

$\displaystyle = \frac{1}{2} \times 9 \times \frac{1}{5}$

$\displaystyle = \frac{9}{10}$

Answer: (i) $200a^{5}$ (ii) $\displaystyle \frac{9}{10}$