Hard GCSE Fractional indices Questions

Challenging, exam-style GCSE Fractional indices questions with worked solutions. Stretch yourself on the hardest fraction base, non-unit fractional index, negative fractional index, equations with fractional indices problems.

fraction basenon-unit fractional indexnegative fractional indexequations with fractional indiceschange of basefractional index equation
GCSE Higher34 questionsStep-by-step solutions
Question 1
5 markschallenging
Write x×x3\sqrt{x} \times \sqrt[3]{x} as a single power of xx.
Show worked solution

Worked solution

  1. Understand the goal

    x×x3=x?\sqrt{x} \times \sqrt[3]{x} = x^?

    Both factors are roots of x. Writing each as a power lets the product law combine them.

  2. Write the square root as a power

    x=x12\sqrt{x} = x^{\frac{1}{2}}

    A square root is the power one half.

  3. Write the cube root as a power

    x3=x13\sqrt[3]{x} = x^{\frac{1}{3}}

    A cube root is the power one third.

  4. Rewrite the product

    x12×x13x^{\frac{1}{2}} \times x^{\frac{1}{3}}

    The expression is now a product of two powers of x.

  5. Recall the product law

    am×an=am+na^m \times a^n = a^{m+n}

    Multiplying powers of the same base means adding the indices.

  6. Set up the index addition

    12+13\frac{1}{2} + \frac{1}{3}

    Add one half and one third — a fraction addition with different denominators.

  7. Find a common denominator

    12=36,13=26\frac{1}{2} = \frac{3}{6},\quad \frac{1}{3} = \frac{2}{6}

    Sixths work for both fractions.

  8. Add the fractions

    36+26=56\frac{3}{6} + \frac{2}{6} = \frac{5}{6}

    Three sixths plus two sixths is five sixths.

  9. Write the single power

    x×x3=x56\sqrt{x} \times \sqrt[3]{x} = x^{\frac{5}{6}}

    So the product is x to the power five sixths.

  10. Interpret the result

    x56=x56x^{\frac{5}{6}} = \sqrt[6]{x^5}

    Five sixths means the sixth root of x to the 5 — a single root, as required.

  11. Check with a value: set x=64x = 64

    64=8,643=4\sqrt{64} = 8,\quad \sqrt[3]{64} = 4

    Test with x=64x = 64: the square root is 8 and the cube root is 4.

  12. Multiply the check values

    8×4=328 \times 4 = 32

    The product is 32.

  13. Evaluate the formula at x=64x = 64

    6456=(646)5=25=3264^{\frac{5}{6}} = (\sqrt[6]{64})^5 = 2^5 = 32

    The sixth root of 64 is 2, and 2 to the 5 is 32 — the same value. ✓

  14. Watch for the common error

    12+13≠15\frac{1}{2} + \frac{1}{3} \ne \frac{1}{5}

    Do not add tops and bottoms separately — a common fractions slip. The correct sum is five sixths.

  15. State the answer

    x56x^{\frac{5}{6}}

    As a single power, the product is x to the power five sixths.

Answer
x56x^{\frac{5}{6}}
Question 2
6 markschallenging
Work out the value of (8116)−34\left(\frac{81}{16}\right)^{-\frac{3}{4}}.
Show worked solution

Worked solution

  1. Read the index carefully

    (8116)−34\left(\frac{81}{16}\right)^{-\frac{3}{4}}

    Three jobs again: the minus flips the fraction, the 4 takes a fourth root, and the 3 cubes the result.

  2. Deal with the negative sign first

    (8116)−34=(1681)34\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{16}{81}\right)^{\frac{3}{4}}

    A negative index on a fraction flips it over.

  3. Recall the fraction-base rule

    (ab)mn=amnbmn\left(\frac{a}{b}\right)^{\frac{m}{n}} = \frac{a^{\frac{m}{n}}}{b^{\frac{m}{n}}}

    Apply the power to top and bottom separately.

  4. Split the fraction

    (1681)34=16348134\left(\frac{16}{81}\right)^{\frac{3}{4}} = \frac{16^{\frac{3}{4}}}{81^{\frac{3}{4}}}

    Two fractional powers to evaluate.

  5. Take the fourth root of 16

    164=2\sqrt[4]{16} = 2

    2×2×2×2=162 \times 2 \times 2 \times 2 = 16, so the fourth root of 1616 is 22.

  6. Cube it

    1634=23=816^{\frac{3}{4}} = 2^3 = 8

    The numerator becomes 8.

  7. Take the fourth root of 81

    814=3\sqrt[4]{81} = 3

    3×3×3×3=813 \times 3 \times 3 \times 3 = 81, so the fourth root of 8181 is 33.

  8. Cube it

    8134=33=2781^{\frac{3}{4}} = 3^3 = 27

    The denominator becomes 27.

  9. Combine the parts

    (8116)−34=827\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \frac{8}{27}

    Put the evaluated top over the evaluated bottom.

  10. Check the flip direction

    8116>1 ⇒ answer<1\frac{81}{16} > 1 \ \Rightarrow\ \text{answer} < 1

    A number above 1 to a negative power gives a number below 1, and eight twenty-sevenths is below 1. ✓

  11. Shortcut view

    16814=23,(23)3=827\sqrt[4]{\frac{16}{81}} = \frac{2}{3},\quad \left(\frac{2}{3}\right)^3 = \frac{8}{27}

    Equivalently: fourth root of the whole fraction is two thirds, then cubing gives eight twenty-sevenths.

  12. Check by inverting the answer

    (827)−1=278=(32)3\left(\frac{8}{27}\right)^{-1} = \frac{27}{8} = \left(\frac{3}{2}\right)^3

    Un-flipping gives twenty-seven eighths, which is three halves cubed — consistent with the fourth root of 81 over 16 being three halves.

  13. Watch for the common error

    do not cube before rooting\text{do not cube before rooting}

    Cubing 16 and 81 first gives huge numbers (4096 and 531441) — root first keeps everything small.

  14. Reflect on the order used

    flip→root→power\text{flip} \to \text{root} \to \text{power}

    Flip, root, power — the same safe order as every negative fractional index question.

  15. State the answer

    827\frac{8}{27}

    The value is eight twenty-sevenths.

Answer
827\frac{8}{27}
Question 3
6 markschallenging
Solve 16x×2=324516^x \times 2 = 32^{\frac{4}{5}}.
Show worked solution

Worked solution

  1. Look at the structure of the equation

    16x×2=324516^x \times 2 = 32^{\frac{4}{5}}

    16, 2 and 32 are all powers of 2, so the whole equation can be written in base 2.

  2. Evaluate the right side first: fifth root

    325=2\sqrt[5]{32} = 2

    For 32 to the power four fifths, take the fifth root first: 2.

  3. Raise to the power 4

    3245=24=1632^{\frac{4}{5}} = 2^4 = 16

    Then apply the power on top: 2 to the 4 is 16. The right side is exactly 16.

  4. Rewrite the equation

    16x×2=1616^x \times 2 = 16

    The equation is now much simpler.

  5. Write 16 as a power of 2

    16=2416 = 2^4

    Convert the remaining numbers to base 2.

  6. Substitute into the left side

    (24)x×21=24(2^4)^x \times 2^1 = 2^4

    Replace 16 with 2 to the 4, and write the lone 2 as 2 to the 1.

  7. Recall the power-of-a-power law

    (am)n=amn(a^m)^n = a^{mn}

    Multiply the indices in the bracket.

  8. Simplify the bracket

    (24)x=24x(2^4)^x = 2^{4x}

    4 times x gives 4x.

  9. Recall the product law

    am×an=am+na^m \times a^n = a^{m+n}

    Multiplying powers of the same base means adding the indices.

  10. Combine the left side

    24x×21=24x+12^{4x} \times 2^1 = 2^{4x+1}

    Add the indices 4x and 1.

  11. Equate the indices

    4x+1=44x + 1 = 4

    Equal powers of 2 must have equal indices, giving a two-step equation.

  12. Subtract 1 from both sides

    4x=34x = 3

    Remove the constant term first.

  13. Solve for x

    x=34x = \frac{3}{4}

    Divide both sides by 4: x is three quarters.

  14. Check the solution

    1634×2=8×2=1616^{\frac{3}{4}} \times 2 = 8 \times 2 = 16

    16 to the three quarters is 8 (fourth root 2, cubed), and 8 times 2 is 16, matching the right side. ✓

  15. State the answer

    x=34x = \frac{3}{4}

    The solution is x equals three quarters.

Answer
x=34x = \frac{3}{4}
Question 4
5 markschallenging
Without using a calculator, estimate the value of 101210^{\frac{1}{2}} to one decimal place. You must show your working.
Show worked solution

Worked solution

  1. Understand the goal

    1012=1010^{\frac{1}{2}} = \sqrt{10}

    The power one half is the square root, so we must estimate root 10 without a calculator.

  2. Bracket with square numbers

    9<10<169 < 10 < 16

    10 sits between the square numbers 9 and 16.

  3. Take roots of the bracket

    3<10<43 < \sqrt{10} < 4

    So root 10 is between 3 and 4 — and much closer to 3, because 10 is close to 9.

  4. Try 3.1

    3.12=9.613.1^2 = 9.61

    Test values one decimal place at a time. 3.1 squared is 9.61, still below 10.

  5. Try 3.2

    3.22=10.243.2^2 = 10.24

    3.2 squared is 10.24, which is past 10.

  6. Narrow the bracket

    3.1<10<3.23.1 < \sqrt{10} < 3.2

    Root 10 is trapped between 3.1 and 3.2.

  7. Decide which end is closer

    10−9.61=0.39,10.24−10=0.2410 - 9.61 = 0.39,\quad 10.24 - 10 = 0.24

    10 is nearer to 10.24 than to 9.61, so root 10 is nearer to 3.2.

  8. Test the midpoint region

    3.162=9.98563.16^2 = 9.9856

    Trying 3.16: its square is just below 10, so root 10 is a little above 3.16.

  9. Confirm with 3.17

    3.172=10.04893.17^2 = 10.0489

    3.17 squared is just past 10, so root 10 lies between 3.16 and 3.17.

  10. Round to one decimal place

    10≈3.2\sqrt{10} \approx 3.2

    Everything between 3.16 and 3.17 rounds to 3.2 at one decimal place.

  11. State the estimate

    1012≈3.210^{\frac{1}{2}} \approx 3.2

    So 10 to the power one half is approximately 3.2.

  12. Sense check against a known root

    9=3\sqrt{9} = 3

    Root 10 should be slightly more than root 9, and 3.2 is slightly more than 3 — consistent.

  13. Watch for the common error

    1012≠510^{\frac{1}{2}} \ne 5

    The power one half means square root, not halving — 10 divided by 2 would wrongly give 5.

  14. Reflect on the method

    bracket→trial squares→round\text{bracket} \to \text{trial squares} \to \text{round}

    Bracketing between known squares and refining one decimal at a time is the standard no-calculator method.

  15. State the answer

    10≈3.2\sqrt{10} \approx 3.2

    To one decimal place, the value is about 3.2.

Answer
1012=10≈3.210^{\frac{1}{2}} = \sqrt{10} \approx 3.2
Question 5
6 markschallenging
Find the smallest integer nn such that 2n2>1002^{\frac{n}{2}} > 100.
Show worked solution

Worked solution

  1. Understand the inequality

    2n2>1002^{\frac{n}{2}} > 100

    We need the smallest whole number n that pushes 2 to the power n over 2 past 100. The fraction in the index is the awkward part.

  2. Plan: remove the fraction

    (2n2)2=2n\left(2^{\frac{n}{2}}\right)^2 = 2^n

    Squaring turns the index n over 2 into n. Squaring both sides of the inequality keeps the order because both sides are positive.

  3. Square both sides

    2n>10022^n > 100^2

    The inequality becomes 2 to the n greater than 100 squared.

  4. Evaluate 100 squared

    1002=10000100^2 = 10000

    So we need the smallest n with 2 to the n above ten thousand.

  5. Recall a landmark power of 2

    210=10242^{10} = 1024

    2 to the power 10 is 1024 — a value worth remembering.

  6. Build up from the landmark

    211=2048, 212=40962^{11} = 2048,\ 2^{12} = 4096

    Keep doubling from 1024.

  7. Continue doubling

    213=81922^{13} = 8192

    8192 is still below 10000.

  8. One more doubling

    214=163842^{14} = 16384

    16384 is past 10000 — the first power of 2 to get there.

  9. Bracket the answer

    213=8192<10000<16384=2142^{13} = 8192 < 10000 < 16384 = 2^{14}

    So the smallest n with 2 to the n above 10000 is 14.

  10. Translate back to the original

    n=14: 2142=27n = 14:\ 2^{\frac{14}{2}} = 2^7

    Check in the original inequality: n=14n = 14 gives 2 to the power 7.

  11. Evaluate the check

    27=128>1002^7 = 128 > 100

    128 beats 100, so n=14n = 14 works. ✓

  12. Check n=13n = 13 fails

    2132=213=81922^{\frac{13}{2}} = \sqrt{2^{13}} = \sqrt{8192}

    For n=13n = 13 the value is the square root of 8192.

  13. Estimate the root

    8192≈90.5<100\sqrt{8192} \approx 90.5 < 100

    90 squared is 8100 and 91 squared is 8281, so root 8192 is between 90 and 91 — below 100. So n=13n = 13 fails.

  14. Confirm minimality

    n=13 fails, n=14 worksn = 13 \text{ fails},\ n = 14 \text{ works}

    Powers of 2 increase with n, so every n below 13 fails too. The smallest winner is 14.

  15. State the answer

    n=14n = 14

    The smallest integer n is 14.

Answer
n=14n = 14

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