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AFCAT Numerical Ability · Indices and Surds

Fractional Powers, Roots as Powers, and Equations in Powers

Read a fractional or decimal index as a root and a power, write every root as an index, and solve an equation by putting both sides on one base.

Why this matters

This is the HARD end of the chapter's index questions: about two in five here are HARD, against a handful on the laws page. They are decimal powers such as a power of 0.25 or 1.5, products of two numbers raised to different decimals, and nested roots. Each one falls to the same two moves: write the base as a power of a prime, and write the index as a fraction.

Concept 1 of 4: A root is a fractional power

a1/2a^{1/2} multiplied by itself is a1/2+1/2=aa^{1/2 + 1/2} = a, so a1/2a^{1/2} must be the square root. In the same way a1/3a^{1/3} is the cube root and a1/na^{1/n} is the nnth root. Once a root is a power, the laws of exponents handle it.

Definition

  • an=a1/n\sqrt[n]{a} = a^{1/n}, and a=a1/2\sqrt{a} = a^{1/2}.
  • To find a1/na^{1/n}, write aa as a perfect nnth power: 32=2532 = 2^5, so 321/5=232^{1/5} = 2.
  • A root raised to a power: (an)k=ak/n\left(\sqrt[n]{a}\right)^k = a^{k/n}.
  • A root of a root multiplies the indices: a3=a1/6\sqrt[3]{\sqrt{a}} = a^{1/6}.

Root as an index

an=a1/n,anm=a1/(mn)\sqrt[n]{a} = a^{1/n}, \qquad \sqrt[m]{\sqrt[n]{a}} = a^{1/(mn)}

Worked example

Find 1251/3125^{1/3} and 40963\sqrt[3]{\sqrt{4096}}.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 1 · Indices and Surds · Fractional Powers and Roots

Simplify: 814\sqrt[4]{81}

A half power is a square root, not half

361/2=636^{1/2} = 6, not 1818. The index 12\tfrac 12 asks for the number that squares to 36.

Nested roots multiply their indices

x3=x1/6\sqrt[3]{\sqrt{x}} = x^{1/6}, not x1/5x^{1/5}. A root of a root is a power of a power, so the indices multiply.

Concept 2 of 4: Powers like m/n and decimal powers

am/na^{m/n} is two steps: the nnth root, then the mmth power. Take the root first and the numbers stay small. A decimal index is only a fraction in disguise, so turn it into one before you start.

Definition

  • am/n=(an)ma^{m/n} = \left(\sqrt[n]{a}\right)^m.
  • Decimals to fractions: 0.25=140.25 = \tfrac 14, 0.5=120.5 = \tfrac 12, 0.75=340.75 = \tfrac 34, 1.5=321.5 = \tfrac 32, 2.5=522.5 = \tfrac 52, 0.6=350.6 = \tfrac 35.
  • A whole-plus-half index leaves a square root: a5/2=a2aa^{5/2} = a^2 \sqrt a.
  • Two different bases with awkward decimals: rewrite both on one prime base, then add the indices. 90.3×270.89^{0.3} \times 27^{0.8} is 30.6×32.43^{0.6} \times 3^{2.4}.
  • A sum such as am/n+bp/qa^{m/n} + b^{p/q}: find each power on its own, then add.

Fractional index

am/n=(an)m=amna^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}

Worked example

Find 320.632^{0.6}.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 2 · Indices and Surds · Fractional Powers and Roots

Evaluate: (256)0.16×(16)0.18(256)^{0.16} \times (16)^{0.18}

A decimal index is a fraction, not a divisor

x0.25x^{0.25} is x1/4x^{1/4}, the fourth root of xx. It is not xx divided by 4, and 6250.25=5625^{0.25} = 5, not 156.25156.25.

Same index: multiply bases. Different indices: one base

80.2×40.2=320.2=28^{0.2} \times 4^{0.2} = 32^{0.2} = 2, because the indices match. With 80.2×40.38^{0.2} \times 4^{0.3} they do not, so rewrite as 20.6×20.6=21.22^{0.6} \times 2^{0.6} = 2^{1.2}.

Root first keeps the numbers small

10002/31000^{2/3}: the cube root is 10 and its square is 100. Squaring first means a cube root of a million, which is the same answer the long way.

Concept 3 of 4: Fractional indices on letters

The laws do not care whether the index is a whole number. x5/6x1/3\dfrac{x^{5/6}}{x^{1/3}} is still top index minus bottom index; the only extra work is subtracting fractions over a common denominator.

Definition

  • Write every root as an index first: a3=a3/2\sqrt{a^3} = a^{3/2} and (a)3=a3/2(\sqrt a)^3 = a^{3/2} as well.
  • Then use the laws: add for a product, subtract for a quotient, multiply for a power of a power.
  • Bring fractions to one denominator before adding or subtracting them: 56−13=56−26=12\tfrac 56 - \tfrac 13 = \tfrac 56 - \tfrac 26 = \tfrac 12.
  • A letter whose indices cancel to 0 becomes 1.

Quotient with fractional indices

xa/bxc/d=xab−cd\dfrac{x^{a/b}}{x^{c/d}} = x^{\frac ab - \frac cd}

Worked example

Simplify p5/6 q4/5p1/3 q2/5\dfrac{p^{5/6}\, q^{4/5}}{p^{1/3}\, q^{2/5}}.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 3 · Indices and Surds · Fractional Powers and Roots

Simplify: x1/2×y1/3x1/4×y1/6\frac{x^{1/2} \times y^{1/3}}{x^{1/4} \times y^{1/6}}

The root of a cube is a three-halves power

a3=a3/2\sqrt{a^3} = a^{3/2}, not a3/2a^{3}/2 and not a6a^{6}. The root divides the index by 2.

Subtract fractions over one denominator

45−13\tfrac 45 - \tfrac 13 is 12−515=715\tfrac{12 - 5}{15} = \tfrac{7}{15}. Subtracting tops and bottoms separately gives 32\tfrac 32, which is wrong.

Concept 4 of 4: Equations in powers and in square roots

If 2x=252^x = 2^5, then x=5x = 5: equal powers of the same base have equal indices. So rewrite both sides on one base and compare indices. An equation with a square root is the other way round: get the root alone, then square it away.

Definition

  • Unknown in the index: write both sides as powers of one prime, then equate indices. A square root on the right is a half power: 93=32.59\sqrt 3 = 3^{2.5}.
  • Unknown under a root: move everything else to the other side first, then square both sides.
  • Put the answer back into the original equation. Squaring can create a value that does not fit.

Equal powers of one base

ax=ak  ⇒  x=k(a>0, a≠1)a^{x} = a^{k} \;\Rightarrow\; x = k \quad (a > 0,\ a \ne 1)

Worked example

Solve 4x+1=324^{x+1} = 32.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionHard

Example 4 · Indices and Surds · Fractional Powers and Roots

Solve: 2x=822^x = 8\sqrt{2}

Isolate the root before squaring

In x+1+2=5\sqrt{x + 1} + 2 = 5, first get x+1=3\sqrt{x + 1} = 3, so x=8x = 8. Squaring the left side as it stands does not remove the root.

Match the bases before matching indices

4x=84^x = 8 does not give x=2x = 2. Write it as 22x=232^{2x} = 2^3, so x=1.5x = 1.5.

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