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Topic 1.10 · HL only

Simplifying expressions (with rational exponents): notes and practice questions

Summary
  • Exponent (Index): A power a base is raised to.
  • Rational Exponent: A fractional power where the denominator is a root and the numerator is a standard power.
  • Index laws apply only when bases are identical.
  • Laws of Indices (must be remembered):
  • Multiplication: xm×xn=xm+nx^m \times x^n = x^{m+n}
  • Division: xm÷xn=xm−nx^m \div x^n = x^{m-n}
  • Power of a Power: (xm)n=xmn(x^m)^n = x^{mn}
  • Product to a Power: (xy)m=xmym(xy)^m = x^m y^m
  • Quotient to a Power: (xy)m=xmym\left(\frac{x}{y}\right)^m = \frac{x^m}{y^m}
  • Power of One: x1=xx^1 = x
  • Power of Zero: x0=1x^0 = 1
  • Negative Exponent: x−m=1xmx^{-m} = \frac{1}{x^m}
  • Rational Exponent (Root): x1n=xnx^{\frac{1}{n}} = \sqrt[n]{x}
  • Rational Exponent (General): xmn=xmn=(xn)mx^{\frac{m}{n}} = \sqrt[n]{x^m} = (\sqrt[n]{x})^m
  • Simplification methods:
  • Apply laws step-by-step.
  • Separate numerical coefficients and algebraic variables.
  • Change bases to match before applying laws (e.g., 94=(32)4=389^4 = (3^2)^4 = 3^8).
  • Handle negative exponents as a final step by rewriting as positive reciprocals.
  • GDC Tips:
  • Evaluate numerical rational exponents directly using brackets: `base^(numerator/denominator)`.
  • Check algebraic simplification by graphing original (y1y_1) and simplified (y2y_2) expressions; they should perfectly overlap.

How it is examined

A one or two mark step inside something larger, most often a power model f(x)=axnf(x) = ax^n where nn is a fraction, or a derivative in AHL 5.9 where the power rule is stated for n∈Qn \in \mathbb{Q}. Rational exponents are why AHL 5.9 can differentiate xnx^n for fractional nn while SL 5.3 cannot.

Key ideas

Simplify expressions, both numerically and algebraically, involving rational exponents.

Linking questions

  • The guide lists no connections for AHL 1.10.

Practice questions

6 questions · 4 medium · 2 hard
Showing 6 of 6

Question 1

MediumPaper 1 · calculator9 marks
(a)

A landscape architect is designing a modular planter box for a new urban garden. The planter box has a base and top that are identical sectors of a circle, each with radius rr cm and angle θ\theta radians. The height of the planter box is h=2h = 2 cm. The total length of metal frame used for all edges of the planter box is L=20L = 20 cm. This includes two circular arcs, four radial edges for the top and bottom sectors, and three vertical connecting edges.

(a) Show that r=72+θr = \frac{7}{2+\theta}.

[2]
(b)(i)

(b) The planter box is designed to hold soil, enclosing a volume, VV.

(i) Find an expression for VV in terms of θ\theta.

[2]
(b)(ii)

(ii) Find the expression for dVdθ\frac{dV}{d\theta}.

[3]
(b)(iii)

(iii) Solve algebraically dVdθ=0\frac{dV}{d\theta} = 0 to find the value of θ\theta that will maximize the volume, VV.

[2]

Question 2

HardPaper 1 · calculator10 marks
(a)

A mathematical model for a physical phenomenon involves the expression (1u+2)2\left(\frac{1}{u} + 2\right)^2.

(a) Expand (1u+2)2\left(\frac{1}{u} + 2\right)^2.

[2]
(b)

The rate of change of a certain quantity is given by f′(x)=1(x+1)2+4x+1+4f'(x) = \frac{1}{(x+1)^2} + \frac{4}{x+1} + 4.

(b) Find the indefinite integral ∫(1(x+1)2+4x+1+4)dx\int \left( \frac{1}{(x+1)^2} + \frac{4}{x+1} + 4 \right) dx.

[4]
(c)

A designer is creating a custom-shaped container. The cross-sectional profile of the container is defined by the curve y=1x+1+2y = \frac{1}{x+1} + 2 for 1≤x≤31 \le x \le 3. The container is formed by rotating this region 2π2\pi about the x-axis.

(c) Calculate the volume of the solid formed. Give your answer in the form π4(a+bln⁡(c))\frac{\pi}{4}(a + b \ln(c) ), where a,b,c∈Za, b, c \in \mathbb{Z}.

[4]

Question 3

MediumPaper 1 · calculator13 marks
(a)(i)

(a) A medical isotope used in diagnostic imaging decays over time. The time tt (in hours) since the isotope was prepared can be modelled by the equation t=−12ln⁡2ln⁡(P)t = \frac{-12}{\ln 2} \ln(P), where PP is the proportion of the isotope remaining.

(i) Calculate the time tt (in hours) when the proportion of the isotope remaining is P=0.5P = 0.5.

[2]
(a)(ii)

(ii) Calculate the time tt (in hours) when the proportion of the isotope remaining is P=0.25P = 0.25.

[2]
(b)

(b) The isotope is considered ineffective if the proportion remaining drops below 0.050.05. Find the time (in hours) when the isotope just becomes ineffective, giving your answer to one decimal place.

[3]
(c)

(c) Express PP in terms of tt.

[3]
(d)

(d) Using your answer from part (c), find the proportion of the isotope remaining after 2424 hours. Give your answer to three significant figures.

[3]

Question 4

HardPaper 2 · calculator11 marks
(a)

(a) A scientist is modeling the growth rate of a specific enzyme in a culture. The instantaneous rate of change is given by the expression:

12x5×5x0×2x3\sqrt{12x^5} \times 5x^0 \times 2x^3

Simplify this expression, assuming x>0x > 0.

[3]
(b)

(b) An astrophysicist is analyzing the energy density fluctuations in a nebula. These fluctuations are described by a ratio of terms involving a variable xx. Simplify the following expression:

(x−3)2(x4)−1\frac{(x^{-3})^2}{(x^4)^{-1}}

Express your answer with a positive exponent.

[3]
(c)

(c) An engineer is designing a new composite material, and the stress distribution in a critical section is modeled by the following expression involving variables xx and yy:

(8x6)2316y8\frac{(8x^6)^{\frac{2}{3}}}{16y^8}

Simplify this expression.

[5]

Question 5

MediumPaper 2 · calculator14 marks
(a)

A company is designing a new line of eco-friendly packaging for artisanal candles. The packaging is in the form of a right prism with a square base. The side length of the square base is xx cm, and the height of the prism is hh cm.

Given that the total external surface area of the box is 960960 cm2^2, show that the height of the box, hh, can be expressed as h=240x−x2h = \frac{240}{x} - \frac{x}{2}.

[2]
(b)

Hence, show that the volume of the box, VV, may be expressed as V=240x−12x3V = 240x - \frac{1}{2}x^3.

[3]
(c)

Sketch the graph of V=240x−12x3V = 240x - \frac{1}{2}x^3, for 0≤x≤200 \le x \le 20.

[2]
(d)

Find an expression for dVdx\frac{dV}{dx}.

[2]
(e)

Find the value of xx which maximizes the volume of the box. Give your answer to three significant figures.

[2]
(f)

Hence, or otherwise, find the maximum possible volume of the box. Give your answer to three significant figures.

[2]
(g)

The company plans to fill these boxes with spherical candles. The design team assumes that they can calculate the exact number of candles in each box by dividing the calculated maximum volume of the box by the volume of a single candle and then rounding down to the nearest integer. Explain why the design team's assumption is incorrect.

[1]

Question 6

MediumPaper 2 · calculator24 marks
(a)(i)

A supermarket display manager is arranging cans in a triangular stack for a promotional event. Each layer of the stack is a row of cans, with the bottom layer having the most cans, and each subsequent layer having fewer cans, forming a triangular shape when viewed from the front. The top layer always has 1 can.

(a) Write down the number of cans in the bottom row of a display with

(i) 4 layers.

[2]
(a)(ii)

(ii) 5 layers.

[2]
(b)(i)

(b) Mayumi notices that the number of cans in the bottom row of the displays forms an arithmetic sequence.

(i) Write down the common difference of this sequence.

[3]
(b)(ii)

(ii) Find an expression for the number of cans in the bottom row of a display with nn layers.

[3]
(c)(i)

(c) The display manager wants to create a stack with 9 layers.

(i) Find the number of cans in the bottom row of this stack.

[3]
(c)(ii)

(ii) Calculate the total number of cans in a stack with 9 layers.

[3]
(d)

(d) Find an expression for the total number of cans in a stack with nn layers, giving your answer in its simplest form.

[2]
(e)

(e) The cans in the stack are either 'regular' or 'promotional'. The number of 'regular' cans in each layer is equal to the layer number (e.g., layer 1 has 1 regular can, layer 2 has 2 regular cans, and so on).

Write down the total number of 'regular' cans in a stack with 4 layers.

[1]
(f)

(f) Find and simplify an expression for the total number of 'regular' cans in a stack with nn layers.

[2]
(g)

(g) The total number of 'promotional' cans in a stack with nn layers is given by the expression n(n−1)2\frac{n(n-1)}{2}.

Using both the given expression and your answer to part (f), find and simplify an expression for the total number of cans (regular and promotional) in a stack with nn layers.

[3]

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What does Simplifying expressions (with rational exponents) cover in IB Maths AI?

Exponent (Index): A power a base is raised to. Rational Exponent: A fractional power where the denominator is a root and the numerator is a standard power. Index laws apply only when bases are identical.

Is Simplifying expressions (with rational exponents) SL or HL?

Simplifying expressions (with rational exponents) is HL only. SL students are not examined on it.

How do I revise Simplifying expressions (with rational exponents) for IB Maths AI?

Start from the core idea: exponent (Index): A power a base is raised to. In the exam: a one or two mark step inside something larger, most often a power model f(x) = ax^n where n is a fraction, or a derivative in AHL 5.9 where the power rule is stated for n ∈ mathbbQ. Rational exponents are why AHL 5.9 can differentiate x^n for fractional n while SL 5.3 cannot. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Simplifying expressions (with rational exponents)?

FourtyFive has 6 Simplifying expressions (with rational exponents) questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

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