Powers, roots and indices

$8 \times 8 = 64$, so $\sqrt{64} = 8$.

Square the whole numbers: $4^2 = 16$, $5^2 = 25$, $6^2 = 36$, $7^2 = 49$.

Any of these works, e.g. 25.

$3^2 = 3 \times 3 = 9$

$7^2 = 7 \times 7 = 49$

Cube numbers are $1, 8, 27, 64, \ldots$

$3^3 = 3 \times 3 \times 3 = 27$, which is in the list.

Squaring means multiplying the number by itself.

$2.5 \times 2.5 = 6.25$

Use the square root key: $\sqrt{17.64} = 4.2$

Check: $4.2 \times 4.2 = 17.64$

$6^5 = 6 \times 6 \times 6 \times 6 \times 6$

$= 7776$

Square numbers: 1, 4, 9, 16, 25, ...

The odd ones are 1, 9, 25, 49, ... so, for example, 9 works ($3^2 = 9$).

$4 \times 4 \times 4 = 64$, so $\sqrt[3]{64} = 4$.

$(-3)^3 = (-3) \times (-3) \times (-3)$

$(-3) \times (-3) = 9$, then $9 \times (-3) = -27$

The reciprocal of a number is 1 divided by it.

$1 \div 0.125 = 8$ (because $0.125 = \frac{1}{8}$).

$1.5 \times 1.5 = 2.25$

So $\sqrt{2.25} = 1.5$

Look for a number that multiplies by itself to give 1.69.

$1.3 \times 1.3 = 1.69$, so $\sqrt{1.69} = 1.3$

(a) $8 \times 8 = 64$, so $\sqrt{64} = 8$

(b) $5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$

Powers of 3: $3^1 = 3$, $3^2 = 9$, $3^3 = 27$, $3^4 = 81$

From the list, the powers of 3 are 9 and 27.

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243$

$1.2 \times 1.2 = 1.44$, so $\sqrt{1.44} = 1.2$

(a) Dividing: subtract the powers, $5 - 2 = 3$, giving $c^3$.

(b) Power of a power: multiply, $4 \times 3 = 12$, giving $d^{12}$.

(a) We need the cube root of 343: $7 \times 7 \times 7 = 343$, so the number is 7.

(b) $4^4 = 4 \times 4 \times 4 \times 4 = 256$

Any non-zero number to the power 0 equals 1, so $10^0 = 1$.

(a) Multiply the numbers: $2 \times 7 = 14$. Add the powers: $x^{3+2} = x^5$, $y^{5+1} = y^6$.

So $14x^5y^6$

(b) Multiply the powers: $2 \times -3 = -6$, so $m^{-6}$

(a)(i) Any non-zero number to the power 0 is 1.

(a)(ii) $5^{-2} = \frac{1}{5^2} = \frac{1}{25}$

(b) Add powers when multiplying, subtract when dividing: $2^{5+4-3} = 2^6$

Inside the bracket add the powers: $2^{-5} \times 2^8 = 2^{3}$

Square by multiplying the power: $(2^3)^2 = 2^6$

(a) Add the powers: $n^{3+5} = n^8$

(b) Subtract the powers: $c^{3-2}d^{4-1} = cd^3$

(c) $5x > 14$, so $x > \frac{14}{5} = 2.8$

(a) Add the powers: $m^{3+4} = m^7$

(b) Cube each part: $5^3 \times n^3 \times (p^3)^3 = 125n^3p^9$

(c) Divide the numbers and subtract the powers: $\frac{32}{4} = 8$, $q^{9-3} = q^6$, $r^{4-1} = r^3$, giving $8q^6r^3$

(a)(i) Any non-zero number to the power 0 is 1.

(ii) $5^{-2} = \frac{1}{5^2} = \frac{1}{25}$

(b) Add the powers when multiplying: $2^5 \times 2^4 = 2^9$. Subtract when dividing: $2^9 \div 2^3 = 2^6$.

Multiplying: add the powers, $-6 + 9 = 3$, so the top is $4^3$.

Dividing by $4^1$: subtract the powers, $4^{3-1} = 4^2 = 16$

A power of $\frac{1}{4}$ means the fourth root, not dividing by 4.

$64^{\frac{1}{4}} = \sqrt[4]{64} = \sqrt{\sqrt{64}} = \sqrt{8} = 2\sqrt{2} \approx 2.83$

$3^7 \times 3^{-2} = 3^{7 + (-2)} = 3^5$

$3^5 \div 3^3 = 3^{5 - 3} = 3^2 = 9$

(a) $25^{\frac{1}{2}} = \sqrt{25} = 5$ and $8^{\frac{1}{3}} = \sqrt[3]{8} = 2$, so the product is $10$.

(b) $\left(\frac{1}{32}\right)^{\frac{1}{5}} = \frac{1}{2}$, then $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$

(a) $10^{150} \times 10^{90} = 10^{240}$

$10^{360} \div 10^{240} = 10^{120}$

$\sqrt{10^{120}} = 10^{60}$

(b) For a power of a power you multiply the indices: $(12^{50})^2 = 12^{100}$. Liam squared 50 instead.

(a) Anything (except 0) to the power 0 is 1, so $7^0 = 1$

(b) $3^6 \times 3^{-6} = 3^0 = 1$, so the value is $3 \times 1 = 3$

(c) $2^{-4} = \frac{1}{2^4} = \frac{1}{16}$

(d) $27^{\frac{1}{3}} = \sqrt[3]{27} = 3$

(a) $\sqrt[4]{81} = 3$ and $\sqrt[4]{10^8} = 10^2$, so the answer is $3 \times 100 = 300$

(b) $64^{\frac{1}{2}} = 8$, and the negative power means the reciprocal: $\frac{1}{8}$

(c) $9^{n-1} = 3^{2(n-1)} = 3^{2n-2}$, so $\frac{3^n}{3^{2n-2}} = 3^{n-(2n-2)} = 3^{2-n}$

(a) $36^{\frac{1}{2}} = \sqrt{36} = 6$

(b) Any non-zero number to the power 0 is 1, so $2^0 = 1$

(c) $27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}} = \frac{1}{(\sqrt[3]{27})^2} = \frac{1}{3^2} = \frac{1}{9}$

(a) $16 = 4^2$, so $\frac{1}{16} = 4^{-2}$

(b) $8^{\frac{5}{3}} = (\sqrt[3]{8})^5 = 2^5 = 32$

$9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$

$32 - 27 = 5$

$27^{\frac{2}{3}} = (\sqrt[3]{27})^2 = 3^2 = 9$

$\left(\frac{1}{2}\right)^{-3} = 2^3 = 8$

$9 + 8 = 17$

The denominator 3 means cube root: $\sqrt[3]{\frac{8}{27}} = \frac{2}{3}$

The numerator 4 means raise to the power 4: $\left(\frac{2}{3}\right)^4 = \frac{16}{81}$

Powers of $x$: $x^{6 \times \frac{1}{n}} = x^3$, so $\frac{6}{n} = 3$ and $n = 2$.

Numbers: $a^{\frac{1}{2}} = 7$, so $a = 7^2 = 49$.

(a) $\left(\frac{16}{81}\right)^{\frac{3}{4}}$: fourth root first gives $\frac{2}{3}$, then cube: $\frac{8}{27}$

(b) $\frac{1}{9} = 3^{-2}$, so $a = -2$

$9\sqrt{3} = 3^2 \times 3^{\frac{1}{2}} = 3^{2.5}$, so $b = 2.5$

$\frac{1}{\sqrt{3}} = 3^{-0.5}$, so $c = -0.5$

$a + b + c = -2 + 2.5 - 0.5 = 0$

The denominator 3 means cube root: $\sqrt[3]{125} = 5$.

The numerator 2 means square: $5^2 = 25$.

$\sqrt[3]{2} = 2^{\frac{1}{3}}$, so $2^x = 2^{n - \frac{1}{3}}$ and $x = n - \frac{1}{3}$

$(\sqrt{2})^5 = (2^{\frac{1}{2}})^5 = 2^{\frac{5}{2}}$, so $y = \frac{5}{2}$

$n - \frac{1}{3} + \frac{5}{2} = 8$

$n = 8 + \frac{1}{3} - \frac{5}{2} = \frac{35}{6}$

$\left(5\frac{4}{9}\right)^{-\frac{1}{2}} = \left(\frac{49}{9}\right)^{-\frac{1}{2}} = \frac{3}{7}$

Numerator: $\frac{3}{7} \times 4\frac{2}{3} = \frac{3}{7} \times \frac{14}{3} = 2$

$2^{-3} = \frac{1}{8}$, so the value is $2 \div \frac{1}{8} = 16$

$9^{-\frac{1}{2}} = 3^{-1}$ and $27^{\frac{1}{4}} = 3^{\frac{3}{4}}$

$3^{-1} = 3^{\frac{3}{4} - (x+1)}$

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

$x = \frac{3}{4}$