Laws of Indices – Number & Algebra – IB Mathematics AA HL

Step 1: Expand the numerator using multiplication.

$(3x^2)(2x^3{y}^2)=3\times 2\times x^{2+3}\times y^2=6x^5{y}^2.$

Step 2: Divide by the denominator $6x^{2}y$.

$\frac{6x^5{y}^2}{6x^{2}y}=\frac{6}{6}\times x^{5-2}\times y^{2-1}=x^{3}y.$

Step 1: Rewrite each factor as a power/fraction if needed.

For $(4x^2{y}^{-4}{)}^3$, apply the law $(abc{)}^n=a^n{b}^n{c}^n$: $(4x^2{y}^{-4}{)}^3=4^3{x}^{2\times 3}{y}^{-4\times 3}=64x^6{y}^{-12}.$

For $(2x^3{y}^{-1}{)}^{-2}$, apply the power: $(2x^3{y}^{-1}{)}^{-2}=2^{-2}{x}^{-6}{y}^2=\frac{y^2}{4x^6}.$

Step 2: Combine them via multiplication.

$64x^6{y}^{-12}\times \frac{y^2}{4x^6}=64\times \frac{1}{4}\times x^6\times x^{-6}\times y^{-12}\times y^2.$

Simplify step by step: $64÷4=16,x^{6+(-6)}=x^0=1,y^{-12+2}=y^{-10}.$

So we get: $16y^{-10}.$

Step 1: Apply the power of a power law: $(x^m{)}^n=x^{mn}$.

$(x^3{)}^4=x^{3\times 4}=x^{12}.$

Step 2: Multiply $x^{12}$ by $x^{-5}$ using $x^m\times x^n=x^{m+n}$.

$x^{12}\times x^{-5}=x^{12+(-5)}=x^7.$

Step 1: Rewrite the fraction to separate numeric and variable factors.

$=\frac{2}{4}\times \frac{x^2}{x^{-3}}\times \frac{y^3}{y^5}.$

Step 2: Simplify each part.

$\frac{2}{4}=\frac{1}{2},\frac{x^2}{x^{-3}}=x^{2-(-3)}=x^5,\frac{y^3}{y^5}=y^{3-5}=y^{-2}.$

Combine them:

$\frac{1}{2}\times x^5\times y^{-2}=\frac{x^5}{2y^2}.$

Step 1: Combine numeric coefficients and then combine like bases.

$2\times 4=8.$
$x^{3/2}\times x^{-1/2}=x^{3/2+(-1/2)}=x^{2/2}=x^1=x.$

Step 2: So the product is:

$8x.$

Step 1: Expand the numerator using $(xy{)}^m=x^m{y}^m$:

$(a^{2}b{)}^3=a^{2\times 3}{b}^{1\times 3}=a^6{b}^3.$

Step 2: Divide by $a^{-1}{b}^4$:

$\frac{a^6{b}^3}{a^{-1}{b}^4}=a^{6-(-1)}{b}^{3-4}=a^7{b}^{-1}.$ which is $a^7\frac{1}{b}.$

Step 1: Inside the parenthesis, note $y^{-2}=\frac{1}{y^2}$. So:

$\frac{x^3}{y^{-2}}=x^3\times y^2.$

Step 2: Now raise to the power $-1$:

$(x^3{y}^2{)}^{-1}=x^{-3}{y}^{-2}.$ Which is $\frac{1}{x^3{y}^2}.$

Step 1: Combine the multiplication first: $x^{3/4}\times x^{5/4}=x^{3/4+5/4}=x^{8/4}=x^2.$

Step 2: Now divide by $x^{2/4}=x^{1/2}.$

So $x^2÷x^{1/2}=x^{2-1/2}=x^{3/2}.$

Step 1: Multiply inside the radical first: $8x^6\times 4x^3=32x^9.$

So the expression is $\sqrt[3]{32x^9}.$

Step 2: Rewrite $32=2^5$, so $32x^9=2^5{x}^9.$

Step 3: Taking the cube root: $\sqrt[3]{2^5{x}^9}=2^{5/3}{x}^{9/3}=2^{5/3}{x}^3.$

Often you might leave $2^{5/3}$ as $2\sqrt[3]{2^2}$ or $2^{1+2/3}$, but in index form $2^{5/3}$ is perfectly acceptable.