Edexcel A level Maths (9MA0) · Pure mathematics › Algebra and functions
Practise Laws of indices. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Write down the value of \(1000^{\frac{1}{3}}\)[1]
(b) Find the value of \(1000^{-\frac{2}{3}}\)[2]
(c) Find the value of \(\left(\frac{4}{25}\right)^{-\frac{3}{2}}\)[2]
Show the answer and mark scheme
(a)Answer: \(10\)
B1 for 10
Worked solution: \(1000^{\frac{1}{3}} = \sqrt[3]{1000} = 10\), since \(10^3 = 1000\).
(b)Answer: \(\frac{1}{100}\)
M1 for a correct first step, e.g. \(\frac{1}{1000^{\frac{2}{3}}}\) or \(\left(1000^{\frac{1}{3}}\right)^{2} = 100\) or \(10^{-2}\)
A1 for \(\frac{1}{100}\) oe (accept 0.01)
Worked solution: \(1000^{-\frac{2}{3}} = \frac{1}{\left(1000^{\frac{1}{3}}\right)^{2}} = \frac{1}{10^{2}} = \frac{1}{100}\).
(c)Answer: \(\frac{125}{8}\)
M1 for inverting the fraction, \(\left(\frac{25}{4}\right)^{\frac{3}{2}}\), or for \(\left(\frac{2}{5}\right)^{-3}\)
A1 for \(\frac{125}{8}\) oe (accept 15.625)
Worked solution: \(\left(\frac{4}{25}\right)^{-\frac{3}{2}} = \left(\frac{25}{4}\right)^{\frac{3}{2}} = \left(\frac{5}{2}\right)^{3} = \frac{125}{8}\).
Question 2Medium6 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Given that \(\frac{\left(x^{3}\sqrt[3]{x}\right)^{3}}{\sqrt[4]{x}} = x^{n}\), where \(x \gt 0\), find the exact value of \(n\).[3]
(b) Solve the equation \(3x^{-\frac{1}{2}} = \frac{3}{5}\)[3]
Show the answer and mark scheme
(a)Answer: \(n = \frac{39}{4}\)
M1 for writing the roots as fractional powers, e.g. \(\sqrt[3]{x} = x^{\frac{1}{3}}\) and \(\sqrt[4]{x} = x^{\frac{1}{4}}\)
M1 for a correct use of the power law on the numerator: \(x^{3 \times \frac{10}{3}} = x^{10}\)
A1 for \(n = \frac{39}{4}\)
Worked solution: \(x^{3}\sqrt[3]{x} = x^{\frac{10}{3}}\), so the numerator is \(x^{10}\). Dividing by \(x^{\frac{1}{4}}\): \(n = 10 - \frac{1}{4} = \frac{39}{4}\).
(b)Answer: \(x = 25\)
M1 for isolating the power: \(x^{-\frac{1}{2}} = \frac{1}{5}\)
M1 for raising both sides to the power \(-2\) (or an equivalent two-step method)
A1 for \(x = 25\)
Worked solution: \(x^{-\frac{1}{2}} = \frac{1}{5}\). \(x = \left(\frac{1}{5}\right)^{-2} = 25\).
Question 3Hard5 marks
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
M1 for writing the equation as a quadratic in \(3^x\), e.g. using \(3^{2x} = (3^x)^2\)
A1 for \(y^{2} + 2y - 3 = 0\) where \(y = 3^x\)
dM1 for solving the quadratic, e.g. by factorising
A1 for \(y = 1\) (or \(y = -3\), rejected)
A1 for \(x = 0\) only, with a reason for rejecting \(3^x = -3\)
Worked solution: Let \(y = 3^x\). Then \(3^{2x} = y^2\) and \(3^{x} = y\). \(y^{2} + 2y - 3 = 0\) has roots \(y = 1\) and \(y = -3\). \(3^x = -3\) has no solutions since \(3^x \gt 0\) for all real \(x\). \(3^x = 1\) gives \(x = 0\).