Indice laws

Number

Questions marked Higher are aimed at the Higher tier.

These are original practice questions, not a past paper. This site is independent and not affiliated with AQA or Pearson.

Simplify $5^{4} \times 5^{3}$. Give the answer as a single power of $5$.

2 marks
Simplify $7^{9} \div 7^{4}$. Give the answer as a single power of $7$.

2 marks
Simplify $(3^{2})^{4}$, and work out the value of $\dfrac{2^{3} \times 2^{4}}{2^{2}}$.

3 marks
Simplify $\dfrac{x^{5} \times x^{4}}{x^{3}}$.

3 marks
Evaluate $16^{1/2}$ and $8^{2/3}$. Higher

4 marks
Evaluate $9^{-1/2}$ and $\left(\dfrac{1}{4}\right)^{-2}$. Higher

4 marks
Simplify $\dfrac{(4x^{3})^{2}}{2x^{4}}$ and $\dfrac{(3a^{2}b)^{3}}{9ab^{2}}$. Higher

5 marks
Work out the value of $\dfrac{9^{3} \times 9^{1/2}}{9^{2}}$. Then find $x$ when $4^{x} = 32$. Higher

6 marks

Mark scheme

Answer. $5^{7}$
- Add the indices.
- $5^{7}$
Answer. $7^{5}$
- Subtract the indices.
- $7^{5}$
Answer. $3^{8}$ and $32$
- $(3^{2})^{4} = 3^{8}$.
- $2^{3} \times 2^{4} \div 2^{2} = 2^{5}$.
- $2^{5} = 32$.
Answer. $x^{6}$
- $x^{9} \div x^{3}$
- $x^{6}$
Answer. $4$ and $4$
- $16^{1/2} = 4$.
- $8^{1/3} = 2$.
- $2^{2} = 4$.
Answer. $\dfrac{1}{3}$ and $16$
- $9^{1/2} = 3$, so $9^{-1/2} = \dfrac{1}{3}$.
- $\left(\dfrac{1}{4}\right)^{-2} = 4^{2} = 16$.
Answer. $8x^{2}$ and $3a^{5}b$
- $(4x^{3})^{2} = 16x^{6}$, so $\dfrac{16x^{6}}{2x^{4}} = 8x^{2}$.
- $(3a^{2}b)^{3} = 27a^{6}b^{3}$.
- $\dfrac{27a^{6}b^{3}}{9ab^{2}} = 3a^{5}b$.
Answer. $27$ and $x = \dfrac{5}{2}$
- The index is $3 + \dfrac{1}{2} - 2 = \dfrac{3}{2}$.
- $9^{3/2} = (9^{1/2})^{3} = 27$.
- $4^{x} = (2^{2})^{x} = 2^{5}$, so $2x = 5$.
- $x = \dfrac{5}{2}$.