Question paper, Set 10

$4^{7}$ means $4$ multiplied by itself 7 times.

$4^{7} = 4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4$

Answer: $4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4$

$3^{4} = 3 \times 3 \times 3 \times 3 = 81$

$4^{3} = 4 \times 4 \times 4 = 64$

$81 > 64$, so $3^{4}$ is greater.

Answer: $3^{4}$, since $81 > 64$

Divide repeatedly by $13$: $2197, 169, 13$.

$2197 = 13 \times 13 \times 13$

$= 13^{3}$

Answer: $2197 = 13^{3}$

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

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

$= 1296$

Answer: $6^{4} = 1296$

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

$(13^{8})^{2} = 13^{8 \times 2}$

$= 13^{16}$

Answer: $13^{16}$

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

$3^{2} \times 7^{2} = (3 \times 7)^{2}$

$= (21)^{2} = 21 \times 21$

$= 441$

Answer: $(21)^{2} = 441$

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

$(2^{5})^{0} = 2^{5 \times 0} = 2^{0} = 1$

Answer: $1$

Divide repeatedly by $8$: $512, 64, 8$, so $512 = 8^{3}$.

$\displaystyle \frac{1}{512} = \frac{1}{8^{3}}$

$= 8^{-3}$ (since $\displaystyle \frac{1}{a^n} = a^{-n}$)

Answer: $8^{-3}$

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

$\displaystyle \left(\frac{1}{8}\right)^{-2} = \left(\frac{8}{1}\right)^{2}$, $\displaystyle \left(\frac{8}{1}\right)^{2} = 8^{2}$

Answer: $8^{2}$

$8$ is even, so $(-1)^{8} = 1$; $13$ is odd, so $(-1)^{13} = -1$.

$(-1)^{8} - (-1)^{13} = 1 - (-1) = 2$

Answer: $2$

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

$(-5)^{5} = -5^{5} = -3125$ ($5$ is odd)

$(-5)^{6} = 5^{6} = 15{,}625$ ($6$ is even)

Answer: $(-5)^{5} = -3125$, $(-5)^{6} = 15{,}625$

Given, $9^{y+1} = 729$

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

$\Rightarrow 9^{y+1} = 9^{3}$

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

$\Rightarrow y+1 = 3$

$\Rightarrow y = 3 - 1$

$\Rightarrow y = 2$

Answer: $y = 2$

Using $\displaystyle \left(\frac{b}{a}\right)^n = \left(\frac{a}{b}\right)^{-n}$: $\displaystyle \left(\frac{2}{7}\right)^{3} = \left(\frac{7}{2}\right)^{-3}$

Using $\displaystyle \frac{1}{a^n} = a^{-n}$: $\displaystyle \frac{1}{4^{5}} = 4^{-5}$

Answer: $\displaystyle \left(\frac{7}{2}\right)^{-3}$, $4^{-5}$

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

$(10b^{2})^{3} = 10^{3} \times (b^{2})^{3}$

$= 1000 \times b^{6} = 1000b^{6}$

Answer: $1000b^{6}$

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

$\displaystyle \frac{(2^{4})^{3} \times 2^{1}}{2^{10}} = \frac{2^{12} \times 2^{1}}{2^{10}} = \frac{2^{13}}{2^{10}}$

Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$: $2^{13-10} = 2^{3}$

Answer: $2^{3}$

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

$4^{-2} \times 3^{-2} = (4 \times 3)^{-2} = 12^{-2}$

$\displaystyle = \frac{1}{12^{2}} = \frac{1}{144}$

Answer: $\displaystyle \frac{1}{144}$

(i) Given, $\displaystyle 6^{x+1} = \frac{1}{36}$

$\displaystyle \frac{1}{36} = \frac{1}{6^{2}} = 6^{-2}$ (dividing $36$ repeatedly by $6$: $36, 6$)

$\Rightarrow 6^{x+1} = 6^{-2}$

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

$\Rightarrow x+1 = -2$

$\Rightarrow x = -2 - 1$

$\Rightarrow x = -3$

(ii) Given, $(2^{2})^{x} \times 2^{3} = 512$

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

$\Rightarrow 2^{2x} \times 2^{3} = 2^{9}$ ($(a^m)^n = a^{m \times n}$)

$\Rightarrow 2^{2x+3} = 2^{9}$ ($a^m \times a^n = a^{m+n}$)

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

$\Rightarrow 2x+3 = 9$

$\Rightarrow 2x = 9 - 3$

$\Rightarrow 2x = 6$

$\displaystyle \Rightarrow x = \frac{6}{2} = 3$

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

(i) $\displaystyle \left(4^{-1} \times 6^{-1}\right) \div 2^{-2} = \left(\frac{1}{4} \times \frac{1}{6}\right) \div \frac{1}{2^{2}}$

$\displaystyle = \frac{1}{24} \div \frac{1}{4}$

$\displaystyle = \frac{1}{24} \times 4 = \frac{1}{6}$

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

$(-3)^{3} + (-6)^{2} = -3^{3} + 6^{2}$

$= -27 + 36$

$= 9$

Answer: (i) $\displaystyle \frac{1}{6}$ (ii) $9$

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

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

$\displaystyle = 2^{\frac{9}{3}} = 2^{3}$

$= 8$

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

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

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

$\displaystyle = \frac{1}{10^{2}} = \frac{1}{100}$

Answer: (i) $8$ (ii) $\displaystyle \frac{1}{100}$

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

$\displaystyle 16^{2} \times \left(\frac{3}{4}\right)^{4} = 256 \times \frac{3^{4}}{4^{4}}$

$\displaystyle = 256 \times \frac{81}{256}$

$= 81$

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

$(-5)^{-7} \div (-5)^{-9} = (-5)^{-7-(-9)} = (-5)^{-7+9} = (-5)^{2}$

$= 25$

Answer: (i) $81$ (ii) $25$