Zero, negative and fractional indices questions with answers

Questions

1. Question 1Zero index

   Evaluate $8^{0}$.

   Show the worked solution

   1. Step 1: Use the zero-index law$8^{0} = 1$
   2. Step 2: Answer$1$
2. Question 2Negative indices

   Evaluate $2^{-1}$.

   Show the worked solution

   1. Step 1: Rewrite with a positive index$2^{-1} = \left(\dfrac{1}{2}\right)^{1}$
   2. Step 2: Answer$\dfrac{1}{2}$
3. Question 3Roots as indices

   Evaluate $\left(\dfrac{125}{216}\right)^{\frac{1}{3}}$.

   Show the worked solution

   1. Step 1: Write the fractional index as a root$\sqrt[3]{\dfrac{125}{216}} = \dfrac{5}{6}$
   2. Step 2: Answer$\dfrac{5}{6}$
4. Question 4Fractional indices

   Evaluate $16^{\frac{3}{4}}$.

   Show the worked solution

   1. Step 1: Write the fractional index as a root$\sqrt[4]{16} = 2$
   2. Step 2: Apply the rest of the index$2^{3} = 8$
   3. Step 3: Answer$8$
5. Question 5Negative fractional indices

   Evaluate $\left(\dfrac{8}{27}\right)^{-\frac{1}{3}}$.

   Show the worked solution

   1. Step 1: Rewrite with a positive index$\left(\dfrac{8}{27}\right)^{-\frac{1}{3}} = \left(\dfrac{27}{8}\right)^{\frac{1}{3}}$
   2. Step 2: Write the fractional index as a root$\sqrt[3]{\dfrac{27}{8}} = \dfrac{3}{2}$
   3. Step 3: Answer$\dfrac{3}{2}$
6. Question 6Simplify with fractional indices

   Simplify $(27a^{-3}b^{6})^{-\frac{2}{3}}$, giving your answer with positive indices.

   Show the worked solution

   1. Step 1: Apply the power to every part of the bracketNumbers: $27^{-\frac{2}{3}} = 3^{-2}$$27^{-\frac{2}{3}} = \dfrac{1}{9}$$a$: $-3 \times \left(-\dfrac{2}{3}\right) = 2$$b$: $6 \times \left(-\dfrac{2}{3}\right) = -4$
   2. Step 2: Write with positive indices$\dfrac{a^{2}}{9b^{4}}$
7. Question 7Index equations

   Solve $2^{x - 2} = 8^{x}$.

   Show the worked solution

   1. Step 1: Write both sides as powers of the same number$2^{x - 2} = 2^{3x}$
   2. Step 2: Same base, so the indices are equal: write that equation$x - 2 = 3x$
   3. Step 3: Collect the x terms on one side and the numbers on the other$2x = -2$
   4. Step 4: Divide both sides by the number in front of x$x = -1$
8. Question 8Index equations: take out the lowest power

   Solve the equation $2^{x + 4} + 2^{x + 1} = 36$.

   Show the worked solution

   1. Step 1: Take out the lowest power, 2^(x + 1), as a common factor$2^{x + 1}\left(2^{3} + 1\right) = 36$$2^{x + 1} \times 9 = 36$
   2. Step 2: Divide both sides by the number in the bracket$2^{x + 1} = \dfrac{36}{9}$$2^{x + 1} = 4$
   3. Step 3: Write 4 as a power of 2: same base, so the indices are equal$2^{x + 1} = 2^{2}$$x + 1 = 2$$x = 1$
9. Question 9Index equations: related bases

   Solve the equation $2^{x} + 2^{x} = 64$.

   Show the worked solution

   1. Step 1: 2 lots of 2^x is 2 × 2^x$2 \times 2^{x} = 64$
   2. Step 2: 2 = 2^1: multiplying, add the indices$2^{x + 1} = 2^{6}$
   3. Step 3: Same base, so the indices are equal$x + 1 = 6$$x = 5$
10. Question 10In terms of a, where a = 3^x

    Given that $a = 5^{x}$, express $\left(\dfrac{1}{5}\right)^{x + 2}$ in terms of $a$.

    Show the worked solution

    1. Step 1: Write 1/5 as a power of 5$\dfrac{1}{5} = 5^{-1}$
    2. Step 2: Multiply the indices, then split the power into two$\left(\dfrac{1}{5}\right)^{x + 2} = 5^{-1(x + 2)}$$= \dfrac{5^{-x}}{5^{2}}$
    3. Step 3: A power of x is a power of 5^x: put in a = 5^x, and work out the number$= \dfrac{(5^{x})^{-1}}{25}$$= \dfrac{1}{25a}$