Indices

Add the indices (including the negatives): $a^{5 + (-2) + (-1)} = a^{2}$. ($a^8$ added the sizes ignoring signs.)

The power applies to every factor: $2^3 \times x^{2\times 3} \times y^{3} = 8x^6y^3$. ($6x^6y^3$ used $2\times 3$ instead of $2^3$.)

A negative index flips the fraction, then apply the power: $\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$. (The negative flips the fraction, it does not make the answer negative.)

Cube-root then square: $27^{\tfrac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$. (3 took only the cube root; 729 squared 27 first.)

The negative gives the reciprocal and the $\tfrac{1}{2}$ gives the square root: $16^{-\tfrac{1}{2}} = \dfrac{1}{\sqrt{16}} = \dfrac{1}{4}$.

$(3a^2b^{-1})^2 = 9a^4b^{-2}$. Multiply by $a^{-1}b^3$: $9a^{4-1}b^{-2+3} = 9a^3b$. ($3a^3b$ forgot to square the 3.)

Divide the coefficients and subtract the indices: $\dfrac{6}{2} = 3$ and $x^{4 - (-1)} = x^5$, giving $3x^5$. (Subtracting $-1$ adds 1 to the index.)

Subtract the indices: $(n+3) - (n-1) = n + 3 - n + 1 = 4$, so the result is $x^4$ (the $n$ terms cancel).

The root is the denominator of the index and the power is the numerator: $\sqrt[3]{x^2} = x^{\tfrac{2}{3}}$. ($x^{\tfrac{3}{2}}$ swapped the numerator and denominator.)

The edge is the cube root of the volume: $(8x^6)^{\tfrac{1}{3}} = 8^{\tfrac{1}{3}} \times (x^6)^{\tfrac{1}{3}} = 2 \times x^2 = 2x^2$ cm. ($8x^2$ did not take the cube root of the 8.)

The side is the square root of the area: $\sqrt{49x^4} = \sqrt{49} \times \sqrt{x^4} = 7 \times x^2 = 7x^2$ cm. ($49x^2$ did not take the square root of the 49.)

Multiply the coefficients and add the powers: $(3 \times 2) \times 10^{4+5} = 6 \times 10^9$. ($6 \times 10^{20}$ multiplied the powers.)

Divide the coefficients and subtract the powers: $\dfrac{8}{4} = 2$ and $10^{7-3} = 10^4$, giving $2 \times 10^4$.

Move the decimal point 4 places right to get $3.4$, so the power is $-4$: $3.4 \times 10^{-4}$. (A number below 1 has a negative power.)

Write both sides with base 2: $4^x = 2^{2x}$ and $32 = 2^5$, so $2x = 5$ and $x = \tfrac{5}{2}$. (Equal bases means equal indices.)

$3^{x+2} = 3^x \times 3^2 = 7 \times 9 = 63$ (no need to find $x$). ($21$ used $7 \times 3$.)

Doubling $5$ times multiplies the amount by $2^5 = 32$: $300 \times 32 = \$9600$. ($\$1500$ used $300 \times 5$ instead of $300 \times 2^5$.)

Write both sides with base 2: $8^x = 2^{3x}$, so $2^{x+3} = 2^{3x}$. Equate the indices: $x + 3 = 3x$, giving $2x = 3$ and $x = \tfrac{3}{2}$.

Evaluate each: $2^{10} = 1024$, $3^6 = 729$, $8^3 = 512$, $4^4 = 256$, $10^2 = 100$. The largest is $2^{10} = 1024$. (A larger base does not always mean a larger value.)

Only $a^m \times a^n = a^{m+n}$ is an index law. The others are traps: $a^m \times b^m = (ab)^m$, indices cannot be added across a sum, $(a+b)^2 = a^2 + 2ab + b^2$, and $a^0 = 1$.

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). ($\sqrt{14}$ multiplied the numbers under the roots instead of adding the squares.)

Top: $x^{\tfrac{1}{2} + \tfrac{3}{2}} = x^2$. Dividing by $x^{-1}$ adds 1: $x^{2 - (-1)} = x^3$.

A reciprocal is written with a negative index: $\dfrac{1}{x^3} = x^{-3}$. ($-x^3$ makes the term negative, which is different.)

Divide: $\dfrac{2 \times 10^6}{5 \times 10^3} = 0.4 \times 10^3 = 400$. (Handle the coefficients and the powers separately.)

Area $= (3 \times 10^2)^2 = 3^2 \times 10^{2 \times 2} = 9 \times 10^4\text{ cm}^2$. ($6 \times 10^4$ used $3 \times 2$ instead of $3^2$, and $9 \times 10^2$ forgot to square the power of ten.)