Laws of indicesAQA A-Level Maths: Subtopic test

1. 1
   A student evaluates numerical powers without a calculator, using the laws of indices.

   (a)

   Find the value of $27^{\frac23}$.

   [1 mark]

   • A$3$
   • B$6$
   • C$9$
   • D$18$

   (b)

   Find the value of $27^{-\frac23}$.

   [1 mark]

   • A$-9$
   • B$\frac19$
   • C$-\frac19$
   • D$\frac1{18}$

   (c)

   Evaluate $\left(\dfrac{27}{8}\right)^{-\frac23}$, giving your answer as a fraction in its simplest form.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Throughout this question, $x>0$.

   (a)

   Simplify $\dfrac{x^{5}\times x^{-2}}{x^{\frac12}}$.

   [1 mark]

   • A$x^{\frac52}$
   • B$x^{\frac32}$
   • C$x^{\frac72}$
   • D$x^{\frac{13}{2}}$

   (b)

   Simplify $\left(8x^{6}\right)^{\frac23}$.

   [1 mark]

   • A$8x^{4}$
   • B$4x^{9}$
   • C$2x^{4}$
   • D$4x^{4}$

   (c)

   Solve the equation $x^{-\frac32}=\dfrac18$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The function $f$ is defined by $f(x)=\dfrac{x^{2}+3x}{\sqrt{x}}$ for $x>0$.

   (a)

   Write $f(x)$ in the form $x^{p}+kx^{q}$, where $p$, $q$ and $k$ are constants to be found.

   [3 marks]

   (b)

   Hence solve the equation $f(x)=4x^{\frac12}$.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   The function $f$ is defined, for $x>0$, by $f(x)=\dfrac{16x^{\frac32}\times x^{-\frac12}}{\left(8x^{6}\right)^{\frac13}}$.

   (a)

   (i) Show that $f(x)=\dfrac8x$.
   (ii) Hence solve $f(x)=x^{-\frac32}$.

   [6 marks]

   (b)

   (i) Solve the equation $f(x)=\sqrt{x}$.
   (ii) Show that $\left[f(x)\right]^{\frac32}\times x^{\frac32}$ is a constant, and find its exact value.

   [6 marks]

   Total for question 4: 12 marks

End of questions