Index Laws and Equations with Indices

A negative power puts only the thing it is attached to on the bottom of a fraction.

1. x^{-3}=\dfrac{1}{x^{3}}
2. So \dfrac{1}{4x^{3}}=\dfrac14\times\dfrac{1}{x^{3}}=\dfrac14x^{-3}
3. 4x^{-3}=\dfrac{4}{x^{3}} and (4x)^{-3}=\dfrac{1}{64x^{3}}, so they are different
4. Answer: \frac14x^{-3}

Deal with the three parts of the power one at a time: the 4 means a fourth root, the 3 means cube, and the minus sign means take the reciprocal.

1. 81^{\frac14}=\sqrt[4]{81}=3
2. 81^{\frac34}=3^{3}=27
3. The negative power gives the reciprocal
4. Answer: \dfrac{1}{27}

Write each side as a single power of p, then the two powers must be equal.

1. Left-hand side: p^{k}\times p^{7}=p^{k+7} (add the powers)
2. Right-hand side: p^{3}\div p^{8}=p^{3-8}=p^{-5} (subtract the powers)
3. k+7=-5
4. Answer: k=-12

Write \sqrt[3]{x} as x^{\frac13} and simplify inside the bracket first.

1. \sqrt[3]{x}=x^{\frac13}
2. Inside the bracket: x^{3}\times x^{\frac13}=x^{\frac{10}{3}} (add the powers)
3. Power of a power: \frac{10}{3}\times\frac35=2 (multiply the powers)
4. Answer: x^{2}

Cube the 2 as well as the x, and write \sqrt{x} as x^{\frac12}.

1. (2x)^3=8x^3 (cube the 2 too)
2. \sqrt{x}=x^{\frac12}, so the top is 8x^3\times x^{\frac12}=8x^{\frac72}
3. Divide the numbers: 8\div16=\frac12
4. Subtract the powers: \frac72-5=-\frac32
5. Answer: \frac12x^{-\frac32}

Write each root as a fractional power of x first: \sqrt[4]{x^{3}}=x^{\frac34}.

1. \sqrt[4]{x^{3}}=x^{\frac34} and \sqrt{x}=x^{\frac12}
2. Bottom: x^{2}\times x^{\frac12}=x^{\frac52}
3. Divide: \frac34-\frac52=\frac34-\frac{10}{4}=-\frac74
4. Answer: x^{-\frac74}

Apply the power -\frac23 to the 125 and to the x^{6} separately.

1. 125^{\frac13}=5, so 125^{\frac23}=25 and 125^{-\frac23}=\dfrac{1}{25}
2. (x^{6})^{-\frac23}=x^{6\times(-\frac23)}=x^{-4} (multiply the powers)
3. Answer: \dfrac{1}{25}x^{-4}

Write 1\frac23 as an improper fraction, then take the reciprocal of both sides.

1. 1\frac23=\dfrac53
2. Reciprocal of both sides: x^{\frac12}=\dfrac35
3. Square both sides: x=\left(\dfrac35\right)^{2}
4. Answer: x=\dfrac{9}{25}

Get the cube root on its own first, then cube both sides.

1. Subtract 7: 2\sqrt[3]{x-5}=-6
2. Divide by 2: \sqrt[3]{x-5}=-3
3. Cube both sides: x-5=(-3)^{3}=-27
4. Answer: x=-22

Write both sides as a power of 5, remembering that \frac{1}{5^x}=5^{-x}.

1. 25=5^2, so 25^{x+1}=5^{2(x+1)}=5^{2x+2}
2. \dfrac{1}{5^x}=5^{-x}
3. Equate the powers: 2x+2=-x
4. 3x=-2
5. Answer: x=-\frac23

Write every root as a fractional power of x; when you multiply, add the powers.

1. \sqrt[3]{x^{2}}=x^{\frac23} and \sqrt[4]{x}=x^{\frac14}
2. Multiply by adding the powers: \frac23+\frac14=\frac{8}{12}+\frac{3}{12}=\frac{11}{12}
3. \sqrt[n]{x^{11}}=x^{\frac{11}{n}}, so \frac{11}{n}=\frac{11}{12}
4. Answer: n=12

Rearrange to get r^{3} on its own, simplify using the index laws, then take the cube root of each power.

1. Multiply by r^{3} and divide by p^{-1}q^{7}: r^{3}=\dfrac{p^{5}q}{p^{-1}q^{7}}
2. p^{5}\div p^{-1}=p^{6} and q\div q^{7}=q^{-6}
3. r^{3}=p^{6}q^{-6}
4. Cube root: divide each power by 3, r=p^{2}q^{-2}
5. Answer: r=\dfrac{p^{2}}{q^{2}}

Write 4 and \sqrt8 as powers of 2, remembering that a square root is a power of \frac12.

1. (2^{x})^{3}=2^{3x} and 4^{x-1}=2^{2(x-1)}=2^{2x-2}
2. Left-hand side: 2^{3x-(2x-2)}=2^{x+2}
3. \sqrt8=(2^{3})^{\frac12}=2^{\frac32}
4. Equate the powers: x+2=\frac32
5. Answer: x=-\frac12

Divide each term on the top by 2x^{2} separately, writing \sqrt{x} as x^{\frac12}.

1. Split the fraction: \dfrac{x^{3}}{2x^{2}}+\dfrac{4x^{\frac12}}{2x^{2}}
2. First term: \dfrac12x^{3-2}=\dfrac12x
3. Second term: 2x^{\frac12-2}=2x^{-\frac32}
4. Answer: \dfrac12x+2x^{-\frac32}

Take out the smallest power of 2 as a common factor on the top and on the bottom.

1. Top: 2^{12}+2^{10}=2^{10}(2^{2}+1)=2^{10}\times5
2. Bottom: 2^{11}-2^{9}=2^{9}(2^{2}-1)=2^{9}\times3
3. \dfrac{2^{10}\times5}{2^{9}\times3}=\dfrac{2\times5}{3}
4. Answer: \dfrac{10}{3}

Take care with the signs: 4^{-x} with x=-\frac32 is 4 to a positive power. Answer part (a) before attempting part (b).

Part (a)

1. y=4^{-(-\frac32)}=4^{\frac32}
2. 4^{\frac12}=2, so 4^{\frac32}=2^{3}
3. Answer: 8

Part (b)

1. \frac14\times16^{x}=8, so 16^{x}=32
2. Write both as powers of 2: 16^{x}=2^{4x} and 32=2^{5}
3. Equate the powers: 4x=5
4. Answer: x=\dfrac54

Substitute x=\frac23 into each function, simplifying inside the bracket before using the power. Answer part (a) before attempting part (b).

Part (a)

1. \dfrac{3x}{4}=\dfrac{3\times\frac23}{4}=\dfrac24=\dfrac12
2. \left(\dfrac12\right)^{-3}=2^{3} (reciprocal, then cube)
3. Answer: 8

Part (b)

1. \mathrm{g}\left(\frac23\right)=\left(p-3\times\frac23\right)^{\frac32}=(p-2)^{\frac32}
2. So (p-2)^{\frac32}=8
3. Raise both sides to the power \frac23: p-2=8^{\frac23}
4. 8^{\frac23}=(\sqrt[3]{8})^{2}=4
5. Answer: p=6

Write every number as a power of 2 (first equation) or a power of 3 (second), including \sqrt3=3^{\frac12}, then equate the powers.

Part (a)

1. 4^x=(2^2)^x=2^{2x} and 32=2^5
2. \dfrac{2^{2x}}{2^y}=2^{2x-y}
3. Equate the powers of 2: 2x-y=5

Part (b)

1. 27^x=(3^3)^x=3^{3x}, so the left side is 3^{3x}\times3^y=3^{3x+y}
2. 9\sqrt3=3^2\times3^{\frac12}=3^{\frac52}
3. Equate the powers of 3: 3x+y=\frac52
4. (or 6x+2y=5)

Part (c)

1. Add the two equations: (2x-y)+(3x+y)=5+\frac52
2. 5x=\frac{15}{2}, so x=\frac32
3. Substitute into 2x-y=5: 3-y=5, so y=-2
4. Check: \dfrac{4^{1.5}}{2^{-2}}=8\times4=32 ✓
5. Answer: x=\frac32,\ y=-2

Write 9 and 27 as powers of 3, and remember that 1=3^{0}. Answer part (a) before attempting part (b).

Part (a)

1. (9^{x})^{x+1}=(3^{2})^{x(x+1)}=3^{2x^{2}+2x}
2. 27^{x+2}=(3^{3})^{x+2}=3^{3x+6}
3. Divide by subtracting the powers: 2x^{2}+2x-(3x+6)
4. Answer: 3^{2x^{2}-x-6}

Part (b)

1. 1=3^{0}, so 3^{2x^{2}-x-6}=3^{0}
2. Equate the powers: 2x^{2}-x-6=0
3. Factorise: (2x+3)(x-2)=0
4. Answer: x=2 or x=-\dfrac32

Write out the bracket twice and multiply every term by every term: there are four products. Answer part (a) before attempting part (b).

Part (a)

1. \left(x^{\frac12}+x^{-\frac12}\right)^{2}=\left(x^{\frac12}+x^{-\frac12}\right)\left(x^{\frac12}+x^{-\frac12}\right)
2. x^{\frac12}\times x^{\frac12}=x and x^{-\frac12}\times x^{-\frac12}=x^{-1}
3. The two middle terms: 2\times x^{\frac12}\times x^{-\frac12}=2x^{0}=2
4. Answer: x+2+x^{-1}

Part (b)

1. From part (a): x+2+\dfrac1x=\dfrac{25}{6}
2. Multiply every term by 6x: 6x^{2}+12x+6=25x
3. 6x^{2}-13x+6=0
4. Factorise: (2x-3)(3x-2)=0
5. Answer: x=\dfrac32 or x=\dfrac23