Exercise 1a (Revision)

$10^5 \times 10^4$

Recall: $a^m \times a^n = a^{m+n}$

$\therefore 10^5 \times 10^4 = 10^{5+4} = 10^9$ 

Answer: $10^9$

$a^3 \times a^9$

Recall: $a^m \times a^n = a^{m+n}$

$\therefore a^3 \times a^9 = a^{3+9} = a^{12}$

Answer: $a^{12}$

$5y \times 4y^4$

First, separate the coefficients from the variables.

$\therefore 5y \times 4y^4$

$= 5 \times y \times 4 \times y^4$

$= (5 \times 4) \times (y \times y^4)$

Recall: $a^m \times a^n = a^{m+n}$

$=20 \times y^{1+4}$

$=20 \times y^5$

$=20y^5$

Answer: $20y^5$

$2^2 \times 2^4$

Recall: $a^m \times a^n = a^{m+n}$

$\therefore 2^2 \times 2^4 = 2^{2+4} = 2^6 = 64$

Answer: $64$

$m^8 \div m^5$

Recall: $a^m \div a^n = a^{m-n}$

$\therefore m^8 \div m^5 = m^{8-5} = m^3$

Answer: $m^3$

$c^7 \div c$

Recall: $a^m \div a^n = a^{m-n}$

$\therefore c^7 \div c = c^{7-1} = c^6$

Answer: $c^6$

$\frac{24x^6}{8x^4}$

Isolate the coefficients from the variables

$=\frac{24 \times x^6}{8 \times x^4}$

$=(\frac{24}{8}) \times (\frac{x^6}{x^4})$

$=3 \times (\frac{x^6}{x^4})$

Recall: $\frac{a^m}{a^n} = a^m \div a^n = a^{m-n}$

Therefore, we can rewrite it as  $3 \times x^{6-4}$

$=3 \times x^{6-4}$

$=3 \times x^2$

$=3x^2$

$\frac{9 \times 10^9}{3 \times 10^3}$

$=\frac{9}{3}\times\frac{10^9}{10^3}$

$=3 \times 10^{9-3}$

$=3 \times 10^6$

$2^0$

Any base to the power of zero is equal to 1

$\therefore 2^0 = 1$

$6 \times z^0$

Any base to the power of zero is 1

$z^0 = 1$

$\therefore 6 \times z^0 = 6 \times 1 = 6$

$4^{-3}$

Recall: $x^{-1} = \frac{1}{x^1}$

$4^{-3} = \frac{1}{4^3}$

$=\frac{1}{64}$

$3x^{-3}$

First, separate the coefficients from the variables.

$3 \times x^{-3}$

Recall: $x^{-1} = \frac{1}{x^1}$

$3x^{-3} = 3 \times x^{-3} = 3 \times \frac{1}{x^3} = \frac{3}{x^3}$

$(1-4)^{-2}$

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

Recall: $x^{-1} = \frac{1}{x^1}$

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

$=\frac{1}{9}$

$(2-3)^{-1}$

$=(-1)^{-1}$

Recall: $x^{-1} = \frac{1}{x^1}$

$=\frac{1}{(-1)^1}$

$=\frac{1}{-1}$

$=-1$

Answer: $-1$

$x^3 \div x^{-5}$

Recall: $a^m \div a^n = a^{m-n}$

$x^3 \div x^{-5}$

$=x^{3-(-5)}$

$=x^{3+5}$

$=x^8$

Answer: $x^8$

$a^-9 \div b^0$

Remember that any base to the power of zero is always 1

$\therefore$ $b^0 = 1$

Thus, $a^{-9} \div b^0 = a^{-9} \div 1 = a^{-9}$

Applying the negative index law

$a^{-9} = \frac{1}{a^9}$

Answer: $\frac{1}{a^9}$ 

$(3x)^{-3}$

The presence of a bracket enclosing $3x$ means that the $^{-3}$ affects both the $3$ and the $x$.

Recall: $a^{-1}=\frac{1}{x^1}$

$\therefore (3x)^{-3} = \frac{1}{(3x)^3}$

$=\frac{1}{(3 \times x)^3}$

$=\frac{1}{3^3 \times x^3}$

$=\frac{1}{27 \times x^3}$

$=\frac{1}{27x^3}$

Answer: $=\frac{1}{27x^3}$

$9a^{-5} \times 4a^6$

Isolate the coefficient from the variables

$=9\times a^{-5} \times 4 \times a^6$

$=(9\times 4) \times (a^{-5} \times a^6)$

$=36 \times a^{-5+6}$

$=36 \times a^1$

$=36a$

Answer: $36a$

$5x^2 \times 4x^0 \times 2x^{-6}$

Isolate the coefficients from the variables.

$=5\times x^2 \times 4 \times x^0 \times 2 \times x^{-6}$

$=(5 \times 4 \times 2) \times (x^2 \times x^0 \times x^{-6})$

$=40 \times x^{2+0+(-6)}$

$=40 \times x^{2+0-6}$

$=40 \times x^{-4}$

$=40 \times \frac{1}{x^4}$ 

$=\frac{40}{x^4}$

$15 \times 10^4 \div (3 \times 10^{-2})$

$=\frac{15 \times 10^4}{3 \times 10^{-2}}$

$=(\frac{15}{3}) \times (\frac{10^4}{10^{-2}})$

$=5 \times 10^{4-(-2)}$

$=5 \times 10^{4+2}$

$=5\times 10^6$  or $5 000000$

Answer: $=5\times 10^6$  or $5 000000$