Question paper, Set 7

$13$ is multiplied by itself 4 times.

$\therefore 13 \times 13 \times 13 \times 13 = 13^{4}$, read as '$13$ power 4'

Answer: $13^{4}$

$3^{7} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$

$= 2187$

Answer: $2187$

Divide repeatedly by $11$: $14{,}641, 1331, 121, 11$.

$14{,}641 = 11 \times 11 \times 11 \times 11$

$= 11^{4}$

Answer: $14{,}641 = 11^{4}$

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

$13^{6} \times 13^{15} = 13^{6+15}$

$= 13^{21}$

Answer: $13^{21}$

$21 = 21^{1}$, Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{21^{6}}{21} = 21^{6-1} = 21^{5}$

Answer: $21^{5}$

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

$5^{2} \times 7^{2} = (5 \times 7)^{2}$

$= (35)^{2} = 35 \times 35$

$= 1225$

Answer: $(35)^{2} = 1225$

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

$\displaystyle \frac{6^{2}}{9^{2}} = \left(\frac{6}{9}\right)^{2}$

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

$\displaystyle = \frac{4}{9}$

Answer: $\displaystyle \frac{4}{9}$

$3^{0} = 1$

$7 \times 3^{0} = 7 \times 1 = 7$

Answer: $7$

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

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

$\displaystyle = \frac{1}{36}$

Answer: $\displaystyle \frac{1}{36}$

$246$ is an even number.

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

$\therefore (-1)^{246} = 1$

Answer: $1$

$5$ is odd, and $(-a)^n = -a^n$ for odd $n$.

$(-10)^{5} = -10^{5} = -1{,}00{,}000$

Answer: $-1{,}00{,}000$

Given, $2^{3n} = 512$

Dividing $512$ repeatedly by $2$: $512, 256, 128, 64, 32, 16, 8, 4, 2$, so $512 = 2^{9}$.

$\Rightarrow 2^{3n} = 2^{9}$

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

$\Rightarrow 3n = 9$

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

Answer: $n = 3$

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

$2^{3x} = (2^{3})^{x}$

$= 8^{x}$

Answer: $8^{x}$

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

$\displaystyle \left(\frac{q^{5}}{2}\right)^{4} = \frac{(q^{5})^{4}}{2^{4}}$

$\displaystyle = \frac{q^{20}}{16}$

Answer: $\displaystyle \frac{q^{20}}{16}$

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

$\displaystyle \frac{2^{8} \times 2^{6}}{2^{9}} = \frac{2^{14}}{2^{9}}$

$= 2^{14-9} = 2^{5}$

Answer: $2^{5}$

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

$3^{-6} \times 3^{8} = 3^{-6+8} = 3^{2}$

$= 9$

Answer: $9$

(i) Given, $2^{2x+1} \div 2^{5} = 8$

Dividing $8$ repeatedly by $2$: $8, 4, 2$, so $8 = 2^{3}$.

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

$\Rightarrow 2^{2x-4} = 2^{3}$

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

$\Rightarrow 2x-4 = 3$

$\Rightarrow 2x = 3 + 4$

$\Rightarrow 2x = 7$

$\displaystyle \Rightarrow x = \frac{7}{2}$

(ii) Given, $(3^{2})^{x} \times 3^{3} = 2187$

Dividing $2187$ repeatedly by $3$: $2187, 729, 243, 81, 27, 9, 3$, so $2187 = 3^{7}$.

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

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

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

$\Rightarrow 2x+3 = 7$

$\Rightarrow 2x = 7 - 3$

$\Rightarrow 2x = 4$

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

Answer: (i) $\displaystyle x = \frac{7}{2}$ (ii) $x = 2$

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

$\displaystyle = \frac{1}{15} \div \frac{1}{25}$

$\displaystyle = \frac{1}{15} \times 25 = \frac{5}{3}$

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

$(-5)^{3} + (-2)^{7} = -5^{3} - 2^{7}$

$= -125 - 128$

$= -253$

Answer: (i) $\displaystyle \frac{5}{3}$ (ii) $-253$

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

$\displaystyle \frac{5^{\frac{11}{3}} \times 5^{-\frac{5}{3}}}{5^{1}} = \frac{5^{\frac{11}{3} + (-\frac{5}{3})}}{5^{1}}$

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

$= 5^{2-1} = 5^{1}$

$= 5$

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

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

$\displaystyle = 5^{\frac{4}{2}} = 5^{2}$

$= 25$

Answer: (i) $5$ (ii) $25$

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

$\displaystyle 27^{2} \times \left(\frac{2}{3}\right)^{6} = 729 \times \frac{2^{6}}{3^{6}}$

$\displaystyle = 729 \times \frac{64}{729}$

$= 64$

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

$(-3)^{-10} \div (-3)^{-13} = (-3)^{-10-(-13)} = (-3)^{-10+13} = (-3)^{3}$

$= -27$

Answer: (i) $64$ (ii) $-27$