Question paper, Set 4

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

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

Answer: $t \times t \times t \times t \times t \times t \times t$

$3^{5} = 3 \times 3 \times 3 \times 3 \times 3 = 243$

$5^{3} = 5 \times 5 \times 5 = 125$

$243 > 125$, so $3^{5}$ is greater.

Answer: $3^{5}$, since $243 > 125$

Divide repeatedly by $7$: $2401, 343, 49, 7$.

$2401 = 7 \times 7 \times 7 \times 7$

$= 7^{4}$

Answer: $2401 = 7^{4}$

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

$4^{3} \times 4^{8} \times 4^{5} = 4^{3+8+5}$

$= 4^{16}$

Answer: $4^{16}$

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

$(11^{6})^{5} = 11^{6 \times 5}$

$= 11^{30}$

Answer: $11^{30}$

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

$3^{4} \times 10^{4} = (3 \times 10)^{4}$

$= (30)^{4} = 30 \times 30 \times 30 \times 30$

$= 8{,}10{,}000$

Answer: $(30)^{4} = 8{,}10{,}000$

$4^{0} = 1$ and $6^{0} = 1$

$4^{0} + 6^{0} = 1 + 1 = 2$

Answer: $2$

Divide repeatedly by $7$: $343, 49, 7$, so $343 = 7^{3}$.

$\displaystyle \frac{1}{343} = \frac{1}{7^{3}}$

$= 7^{-3}$ (since $\displaystyle \frac{1}{a^n} = a^{-n}$)

Answer: $7^{-3}$

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

$\displaystyle \left(\frac{9}{2}\right)^{-5} = \left(\frac{2}{9}\right)^{5}$

Answer: $\displaystyle \left(\frac{2}{9}\right)^{5}$

$18$ is even, so $(-1)^{18} = 1$; $11$ is odd, so $(-1)^{11} = -1$.

$(-1)^{18} \times (-1)^{11} = 1 \times (-1) = -1$

Answer: $-1$

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

$(-7)^{2} = 7^{2} = 49$ ($2$ is even)

$(-7)^{3} = -7^{3} = -343$ ($3$ is odd)

Answer: $(-7)^{2} = 49$, $(-7)^{3} = -343$

Given, $7^{2x+3} = 49$

Dividing $49$ repeatedly by $7$: $49, 7$, so $49 = 7^{2}$.

$\Rightarrow 7^{2x+3} = 7^{2}$

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

$\Rightarrow 2x+3 = 2$

$\Rightarrow 2x = 2 - 3$

$\Rightarrow 2x = -1$

$\displaystyle \Rightarrow x = -\frac{1}{2}$

Answer: $\displaystyle x = -\frac{1}{2}$

Using $1 = a^0$: $1 = 7^{0}$

Using $\displaystyle \frac{1}{a} = a^{-1}$: $\displaystyle \frac{1}{7} = 7^{-1}$

Answer: $1 = 7^{0}$, $\displaystyle \frac{1}{7} = 7^{-1}$

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

$(2x^{2}y^{5})^{3} = 2^{3} \times (x^{2})^{3} \times (y^{5})^{3}$

$= 8x^{6}y^{15}$

Answer: $8x^{6}y^{15}$

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

$\displaystyle \frac{7^{11} \times 7^{5}}{7^{14}} = \frac{7^{16}}{7^{14}}$

$= 7^{16-14} = 7^{2}$

Answer: $7^{2}$

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

$2^{-3} \times 5^{-3} = (2 \times 5)^{-3} = 10^{-3}$

$\displaystyle = \frac{1}{10^{3}} = \frac{1}{1000}$

Answer: $\displaystyle \frac{1}{1000}$

(i) Given, $3^{3x} \div 3^{2} = 81$

Dividing $81$ repeatedly by $3$: $81, 27, 9, 3$, so $81 = 3^{4}$.

$\Rightarrow 3^{3x-2} = 3^{4}$ ($\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

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

$\Rightarrow 3x-2 = 4$

$\Rightarrow 3x = 4 + 2$

$\Rightarrow 3x = 6$

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

(ii) Given, $\displaystyle 2^{x+3} = \frac{1}{64}$

$\displaystyle \frac{1}{64} = \frac{1}{2^{6}} = 2^{-6}$ (dividing $64$ repeatedly by $2$: $64, 32, 16, 8, 4, 2$)

$\Rightarrow 2^{x+3} = 2^{-6}$

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

$\Rightarrow x+3 = -6$

$\Rightarrow x = -6 - 3$

$\Rightarrow x = -9$

Answer: (i) $x = 2$ (ii) $x = -9$

(i) $\displaystyle \left(2^{15} \div 2^{11}\right) \times 2^{3} = \frac{2^{15}}{2^{11}} \times 2^{3}$

$= 2^{15-11} \times 2^{3} = 2^{4} \times 2^{3}$

$= 2^{4+3} = 2^{7}$

$= 128$

(ii) $\displaystyle \left(2^{-1} \times 3^{-1}\right) \div 6^{-2} = \left(\frac{1}{2} \times \frac{1}{3}\right) \div \frac{1}{6^{2}}$

$\displaystyle = \frac{1}{6} \div \frac{1}{36}$

$\displaystyle = \frac{1}{6} \times 36 = 6$

Answer: (i) $128$ (ii) $6$

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

$\displaystyle \frac{2^{\frac{13}{2}} \times 2^{-\frac{3}{2}}}{2^{3}} = \frac{2^{\frac{13}{2} + (-\frac{3}{2})}}{2^{3}}$

$\displaystyle = \frac{2^{\frac{10}{2}}}{2^{3}} = \frac{2^{5}}{2^{3}}$

$= 2^{5-3} = 2^{2}$

$= 4$

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

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

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

$= 27$

Answer: (i) $4$ (ii) $27$

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

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

$= 4y^{6} \times 9y^{2}$

$= 36 \cdot y^{6+2} = 36y^{8}$

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

$\displaystyle 25^{2} \times \left(\frac{2}{5}\right)^{4} = 625 \times \frac{2^{4}}{5^{4}}$

$\displaystyle = 625 \times \frac{16}{625}$

$= 16$

Answer: (i) $36y^{8}$ (ii) $16$