Exponent laws extended

$1$ .
$1$ .
$\dfrac{1}{9}$ .
$\dfrac{1}{100\,000}$ .
$\dfrac{1}{a^2}$ . Method: $a^{3+(-5)} = a^{-2} = \dfrac{1}{a^2}$ .
$\dfrac{1}{m^3}$ . Method: $m^{4-7} = m^{-3}$ .
$x^{-6} = \dfrac{1}{x^6}$ . Method: $(-3) \times 2 = -6$ .
$\dfrac{1}{8a^3}$ . Method: $2^{-3} \times a^{-3} = \dfrac{1}{8} \times \dfrac{1}{a^3}$ .
$16$ . Method: $\left(\dfrac{1}{2}\right)^{-4} = 2^4 = 16$ .
$5.6 \times 10^{-4}$ .

$\dfrac{a^6}{b^5}$ . Method: $a^{5-(-1)} = a^6$ ; $b^{-2-3} = b^{-5}$ .
$\dfrac{27y^3}{x^6}$ . Method: $3^3 x^{-6} y^3 = 27 x^{-6} y^3$ .
$\dfrac{p^9}{q^7}$ . Method: numerator $p^6 q^{-3}$ ; division gives $p^{6-(-3)} q^{-3-4} = p^9 q^{-7}$ .
$a^{-1} + b^{-1} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$ . But $(a+b)^{-1} = \dfrac{1}{5} = 0.2$ . Since $\dfrac{5}{6} \neq \dfrac{1}{5}$ , the two expressions are not equal.
$\dfrac{3y^3}{x^5}$ . Method: $\dfrac{12}{4} = 3$ ; $x^{-3-2} = x^{-5}$ ; $y^{2-(-1)} = y^3$ .
$80$ nm. Method: $8 \times 10^{-8} \text{ m} = 80 \times 10^{-9} \text{ m} = 80$ nm.
$\dfrac{b^{-2}}{2a^3} = \dfrac{1}{2a^3 b^2}$ . Method: flip and apply positive exponent.
$x^2 = \dfrac{1}{9}$ . Since $x^{-2} = 9$ , take the reciprocal: $x^2 = \dfrac{1}{9}$ .

The quotient rule says $\dfrac{a^m}{a^n} = a^{m-n}$ . When $m < n$ , we get $a^{m-n}$ with a negative exponent. But direct cancellation gives $\dfrac{1}{a^{n-m}}$ . For these to be equal, $a^{-(n-m)}$ must equal $\dfrac{1}{a^{n-m}}$ , which means $a^{-k} = \dfrac{1}{a^k}$ .
$\dfrac{25a^4 b^{-6}}{a^{-3} b^6} = 25 a^{4-(-3)} b^{-6-6} = 25 a^7 b^{-12} = \dfrac{25a^7}{b^{12}}$ .
Intensity at distance $3d$ : $I_0 \times (3d)^{-2} = I_0 \times \dfrac{1}{9d^2} = \dfrac{1}{9} \times I_0 d^{-2}$ . The intensity decreases by a factor of $9$ .
Let $x = 1, y = 1$ . Then $(1+1)^{-1} = \dfrac{1}{2}$ , but $1^{-1} + 1^{-1} = 2$ . The error is distributing an exponent over addition — the power-of-a-product rule applies to multiplication, not addition.
$(2 \times 10^3)^{-2} = 2^{-2} \times 10^{-6} = \dfrac{1}{4} \times 10^{-6}$ . Multiply by $5 \times 10^{-1}$ : $\dfrac{5}{4} \times 10^{-7} = 1.25 \times 10^{-7}$ . Divide by $10^{-4}$ : $1.25 \times 10^{-7-(-4)} = 1.25 \times 10^{-3}$ .

$\dfrac{a^{2n} \times a^{-n+1}}{a^{2n-2}} = \dfrac{a^{2n + (-n+1)}}{a^{2n-2}} = \dfrac{a^{n+1}}{a^{2n-2}} = a^{(n+1)-(2n-2)} = a^{3-n}$ .
$(a^m)^{-n}$ : multiply exponents gives $a^{m \times (-n)} = a^{-mn}$ . $(a^{-m})^n$ : multiply exponents gives $a^{(-m) \times n} = a^{-mn}$ . Both equal $a^{-mn}$ .
Fraction remaining after $t$ years: $\left(\dfrac{1}{2}\right)^t = 2^{-t}$ . Need $2^{-t} < 10^{-3}$ , i.e. $2^t > 1000$ . Since $2^{10} = 1024 > 1000$ , after $10$ years less than $\dfrac{1}{1000}$ remains.
$(3x^{-2}y^3)^{-2} = 3^{-2} x^4 y^{-6} = \dfrac{x^4}{9y^6}$ . $(9x^4 y^{-1})^2 = 81 x^8 y^{-2}$ . Product: $\dfrac{81 x^{12}}{9 y^8} = 9 x^{12} y^{-8}$ . Divide by $(xy)^{-4} = x^{-4} y^{-4}$ : $9 x^{12-(-4)} y^{-8-(-4)} = 9 x^{16} y^{-4} = \dfrac{9x^{16}}{y^4}$ .

Tier 1