Indices & Surds

$a^m \times a^n = a^{m+n}$ is the multiplication law and always holds. ($(a^m)^n = a^{mn}$, not $a^{m+n}$; $(a+b)^2 = a^2 + 2ab + b^2$; $\sqrt{a^2} = |a|$, which equals $a$ only when $a \ge 0$.)

Write $6^4 = (2 \times 3)^4 = 2^4 \times 3^4$, so $\dfrac{2^4 \times 3^4}{2^4 \times 3^2} = 3^{4-2} = 3^2 = 9$.

$(a^3 b^{-2})^2 = a^6 b^{-4}$, then $a^6 b^{-4} \times a^{-1} b^3 = a^{6-1} b^{-4+3} = a^5 b^{-1} = \dfrac{a^5}{b}$.

Write both sides with base $2$: $4^{x+1} = 2^{2x+2}$ and $32 = 2^5$, so $2x + 2 = 5$ and $x = \dfrac{3}{2}$.

A negative index flips the fraction: $\left(\dfrac{27}{8}\right)^{\tfrac{2}{3}} = \left(\sqrt[3]{\dfrac{27}{8}}\right)^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.

$2^{x+3} = 2^x \times 2^3 = 5 \times 8 = 40$. ($15$ used $5 \times 3$; an added index means you multiply by $2^3$.)

Flip the fraction and apply the positive index: $\left(\dfrac{2}{3}\right)^{-2} = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$. ($\dfrac{4}{9}$ squared without flipping.)

Cube the bracket, then divide: $(2a^2b)^3 = 8a^6b^3$, so $\dfrac{8a^6b^3}{4a^3b} = 2a^{6-3}b^{3-1} = 2a^3b^2$. ($8a^3b^2$ forgot to divide the coefficient; $2a^6b^2$ forgot to subtract the indices.)

Subtract the indices: $x^{(n+2)-(n-1)} = x^{n+2-n+1} = x^3$ (the $n$ terms cancel).

Use the distance formula: $d = \sqrt{(4-1)^2 + (8-2)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45}$. Then simplify the surd: $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$.

By Pythagoras, $h^2 = (\sqrt{7})^2 + (\sqrt{2})^2 = 7 + 2 = 9$, so $h = \sqrt{9} = 3$ cm. (Squaring a surd removes the root, so the two sides combine to a perfect square.)

A square root is the $\tfrac{1}{2}$ power, then apply the power-of-a-power law: $\sqrt{x^3} = (x^3)^{\tfrac{1}{2}} = x^{3 \times \tfrac{1}{2}} = x^{\tfrac{3}{2}}$. ($x^{\tfrac{2}{3}}$ inverted the fraction; $x^6$ multiplied $3 \times 2$.)

The side length is $\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$ cm, so the perimeter is $4 \times 2\sqrt{3} = 8\sqrt{3}$ cm.

Negative and fractional together: $9^{-\tfrac{1}{2}} = \dfrac{1}{9^{\tfrac{1}{2}}} = \dfrac{1}{\sqrt{9}} = \dfrac{1}{3}$. ($-3$ confused a negative index with a negative value.)

Take out the largest square factor: $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\,\sqrt{2} = 5\sqrt{2}$. ($2\sqrt{5}$ comes from $\sqrt{20}$, not $\sqrt{50}$.)

$\text{distance} = (3 \times 10^8)(2 \times 10^{-3}) = 6 \times 10^{8 + (-3)} = 6 \times 10^5$ m. ($6 \times 10^{11}$ added the indices as $8 + 3$ instead of $8 + (-3)$.)

Area $= \sqrt{18} \times \sqrt{8} = \sqrt{18 \times 8} = \sqrt{144} = 12\text{ cm}^2$. (Multiplying two surds can give a whole number.)

Simplify each surd first: $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{27} = 3\sqrt{3}$. They are like surds, so $2\sqrt{3} + 3\sqrt{3} = 5\sqrt{3}$. ($\sqrt{39}$ wrongly added the numbers under one root.)

The edge is the cube root of the volume: $(27x^9)^{\tfrac{1}{3}} = 27^{\tfrac{1}{3}} \times (x^9)^{\tfrac{1}{3}} = 3 \times x^{9 \times \tfrac{1}{3}} = 3x^3$ cm.

This is a difference of two squares: $(\sqrt{5})^2 - 2^2 = 5 - 4 = 1$. (Multiplying a surd by its conjugate always gives a rational number, which is the key to rationalising denominators.)

Multiply top and bottom by $\sqrt{5}$: $\dfrac{3}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{3\sqrt{5}}{5}$ (since $\sqrt{5} \times \sqrt{5} = 5$). ($\dfrac{3}{5}$ forgot the surd in the numerator.)

$2^{12} = 4096$, $5^5 = 3125$, $3^7 = 2187$, $6^4 = 1296$, $10^3 = 1000$. The largest is $2^{12} = 4096$ (a bigger index can beat a bigger base).

Power of a product with a fractional index: $(8x^6)^{\tfrac{1}{3}} = 8^{\tfrac{1}{3}} \times (x^6)^{\tfrac{1}{3}} = \sqrt[3]{8} \times x^{6 \times \tfrac{1}{3}} = 2x^2$. ($8x^2$ took the cube root of $x^6$ but not of 8.)

Write every term as a power of $3$: $3^x \times 9 = 3^{x+2}$ and $27^{x-1} = 3^{3(x-1)} = 3^{3x-3}$. Equating the indices gives $x + 2 = 3x - 3$, so $5 = 2x$ and $x = \dfrac{5}{2}$.

Multiply top and bottom by the conjugate $\sqrt{3} + 1$: the denominator becomes $(\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2$, so the expression is $\dfrac{\sqrt{3} + 1}{2}$. (Multiplying by the conjugate clears the surd from the denominator.)