Question paper, Set 3

$m$ is multiplied by itself 6 times.

$\therefore m \times m \times m \times m \times m \times m = m^{6}$, read as '$m$ power 6'

Answer: $m^{6}$

$11^{3} = 11 \times 11 \times 11$

$= 1331$

Answer: $1331$

Divide repeatedly by $5$: $3125, 625, 125, 25, 5$.

$3125 = 5 \times 5 \times 5 \times 5 \times 5$

$= 5^{5}$

Answer: $3125 = 5^{5}$

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

$\displaystyle \left(\frac{2}{5}\right)^{3} \times \left(\frac{2}{5}\right)^{6} = \left(\frac{2}{5}\right)^{3+6}$

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

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

$13 = 13^{1}$, Using $\displaystyle \frac{a^m}{a^n} = a^{m-n}$

$\displaystyle \frac{13^{9}}{13} = 13^{9-1} = 13^{8}$

Answer: $13^{8}$

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

$5^{3} \times 6^{3} = (5 \times 6)^{3}$

$= (30)^{3} = 30 \times 30 \times 30$

$= 27{,}000$

Answer: $(30)^{3} = 27{,}000$

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

$\displaystyle \frac{35^{3}}{7^{3}} = \left(\frac{35}{7}\right)^{3}$

$= 5^{3}$

$= 125$

Answer: $125$

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

$\displaystyle \left(\frac{5}{11}\right)^{0} = 1$

Answer: $1$

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

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

$\displaystyle = \frac{1}{64}$

Answer: $\displaystyle \frac{1}{64}$

$84$ is an even number.

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

$\therefore (-1)^{84} = 1$

Answer: $1$

$3$ is odd, and $(-a)^n = -a^n$ for odd $n$.

$(-9)^{3} = -9^{3} = -729$

Answer: $-729$

Given, $5^{3m+1} = 625$

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

$\Rightarrow 5^{3m+1} = 5^{4}$

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

$\Rightarrow 3m+1 = 4$

$\Rightarrow 3m = 4 - 1$

$\Rightarrow 3m = 3$

$\displaystyle \Rightarrow m = \frac{3}{3} = 1$

Answer: $m = 1$

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

$\displaystyle 2^{x-5} = \frac{2^{x}}{2^{5}}$

$\displaystyle = \frac{2^{x}}{32}$

Answer: $\displaystyle \frac{2^{x}}{32}$

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

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

$= 27 \times m^{12} = 27m^{12}$

Answer: $27m^{12}$

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

$(xy^{2})^{3} \times (x^{2}y)^{4} = x^{3}y^{6} \times x^{8}y^{4}$

Using $a^m \times a^n = a^{m+n}$: $= x^{11}y^{10}$

Answer: $x^{11}y^{10}$

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

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

$\displaystyle = \frac{1}{-32} = -\frac{1}{32}$ ($5$ is odd)

Answer: $\displaystyle -\frac{1}{32}$

(i) Given, $7^{2x-1} \times 7^{3} = 2401$

Dividing $2401$ repeatedly by $7$: $2401, 343, 49, 7$, so $2401 = 7^{4}$.

$\Rightarrow 7^{2x-1+3} = 7^{4}$ ($a^m \times a^n = a^{m+n}$)

$\Rightarrow 7^{2x+2} = 7^{4}$

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

$\Rightarrow 2x+2 = 4$

$\Rightarrow 2x = 4 - 2$

$\Rightarrow 2x = 2$

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

(ii) Given, $(-3)^{x+2} \times (-3)^{4} = (-3)^{9}$

$\Rightarrow (-3)^{x+2+4} = (-3)^{9}$ ($a^m \times a^n = a^{m+n}$)

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

$\Rightarrow x+6 = 9$

$\Rightarrow x = 9 - 6$

$\Rightarrow x = 3$

Answer: (i) $x = 1$ (ii) $x = 3$

(i) $\displaystyle \left(5^{0} + 3^{-1}\right) \times 3^{2} = \left(1 + \frac{1}{3}\right) \times 9$

$\displaystyle = \frac{4}{3} \times 9$

$= 12$

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

$(-3)^{4} + (-2)^{5} = 3^{4} - 2^{5}$

$= 81 - 32$

$= 49$

Answer: (i) $12$ (ii) $49$

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

$\displaystyle 2^{\frac{7}{2}} \times 2^{\frac{5}{2}} = 2^{\frac{7}{2} + \frac{5}{2}}$

$\displaystyle = 2^{\frac{12}{2}} = 2^{6}$

$= 64$

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

$\displaystyle 5^{\frac{7}{3}} \div 5^{\frac{1}{3}} = 5^{\frac{7}{3} - \frac{1}{3}}$

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

$= 25$

Answer: (i) $64$ (ii) $25$

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

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

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

$= 3^{1} \times 4^{-2} \times 5^{1}$ ($a^m \times a^n = a^{m+n}$, $\displaystyle \frac{a^m}{a^n} = a^{m-n}$)

$\displaystyle = 3 \times \frac{1}{16} \times 5$

$\displaystyle = \frac{15}{16}$

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

$(-2)^{-15} \div (-2)^{-12} = (-2)^{-15-(-12)} = (-2)^{-15+12} = (-2)^{-3}$

$\displaystyle = \frac{1}{(-2)^{3}} = \frac{1}{-8} = -\frac{1}{8}$

Answer: (i) $\displaystyle \frac{15}{16}$ (ii) $\displaystyle -\frac{1}{8}$