Why does 9 to the power of one half equal 3 and not 4.5?

Because the index 1/2 means square root, not divide by 2. The rule is a to the 1/n equals the nth root of a. So 9 to the 1/2 equals the square root of 9, which is 3, because 3 squared is 9. The answer 4.5 comes from dividing 9 by 2, which is the wrong operation.

What does the denominator of a fractional index tell you?

The denominator tells you which root to take. a to the 1/n means the nth root of a. So a to the 1/2 is the square root, a to the 1/3 is the cube root, a to the 1/4 is the fourth root, and so on. The denominator is never a divisor.

How do you evaluate (5 and 1/16) to the power of 1/4?

First convert the mixed number: 5 and 1/16 equals 81/16. Then apply the fourth root to numerator and denominator separately: the fourth root of 81 is 3 (because 3 to the 4 is 81) and the fourth root of 16 is 2 (because 2 to the 4 is 16). So the answer is 3/2 = 1.5. The trap of dividing by 4 gives 81/64, which is wrong.

GCSE Maths Higher · AQA · Index laws and standard form

Fractional indices mean roots, not division

When students see $9^{1/2}$ , the instinct is to read the index 1/2 as “divide by 2” and write 4.5. That answer is wrong. The correct rule is $a^{1/n} = \sqrt[n]{a}$ : a fractional index means a root, not a division.

So $9^{1/2} = \sqrt{9} = 3$ , because $3^2 = 9$ . The trap answer 4.5 = 9/2 uses the wrong operation entirely. This confusion is one of the most reliable grade 7 to 9 mark-losers on AQA Higher index-law papers.

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How to spot it in your own work

You wrote 4.5 as the value of $9^{1/2}$ , dividing 9 by 2 instead of taking the square root.
You evaluated $(5\tfrac{1}{16})^{1/4}$ by dividing by 4 rather than taking the fourth root of numerator and denominator separately.
Given $7^n = x$ , you wrote $7^{\tfrac{1}{2}n} = \tfrac{x}{2}$ instead of $\sqrt{x}$ .
You applied $a^{m/n}$ by multiplying by m/n rather than taking the nth root first and then raising to the mth power.

An exam question that triggers it

Here is the structure of a typical AQA Higher index-laws question:

Evaluate $\left(5\tfrac{1}{16}\right)^{1/4}$ .

The misconception divides by 4 to get $\frac{81}{64} \approx 1.27$ . The fix: convert to $\frac{81}{16}$ , take the fourth root of numerator and denominator: $\frac{81^{1/4}}{16^{1/4}} = \frac{3}{2} = 1.5$ .

Why students fall for this

Students see a fraction in the index and associate it with the arithmetic operation of multiplication or division by that fraction. The index 1/2 in $9^{1/2}$ looks like “multiply by one half” or “divide by two,” giving 4.5. The root interpretation requires knowing that a fractional exponent encodes a root, not a scalar operation.

A second error pattern is applying the root only to part of the expression. In $(81/16)^{1/4}$ , students sometimes root only the numerator (getting 3/16) while leaving the denominator unchanged. The correct method applies the root to both parts.

Both errors are reinforced by the plausibility of the wrong answer. A student who writes 4.5 for $9^{1/2}$ often has no sense that anything is wrong, which is what makes this a reliable mark-loser at grades 7 to 9.

The fix: Fractional index means root: a^(1/n) = nth root of a

Step 1: identify the denominator. In $a^{m/n}$ , the denominator n gives the root and the numerator m gives the power. For $9^{1/2}$ , n = 2 means square root, m = 1 means no additional power.

Step 2: apply the root. Take the nth root of a. For $9^{1/2}$ : $\sqrt{9} = 3$ because $3^2 = 9$ .

Step 3: apply the numerator power (if any). For $8^{2/3}$ : cube root of 8 is 2 (since $2^3 = 8$ ), then $2^2 = 4$ . So $8^{2/3} = 4$ .

Step 4: sense-check. For a > 1, the square root is smaller than a. If your answer for $9^{1/2}$ is larger than 9 or equals 4.5, recheck whether you divided instead of rooting.

Worked example

Evaluate $9^{1/2}$ .

Rule. $a^{1/2} = \sqrt{a}$ , so
$9^{1/2} = \sqrt{9} = 3$
Trap: $9 \div 2 = 4.5$ , which divides instead of rooting. Check: $3^2 = 9$ .

Evaluate $\left(5\tfrac{1}{16}\right)^{1/4}$ .

Convert. $5\tfrac{1}{16} = \tfrac{80}{16} + \tfrac{1}{16} = \tfrac{81}{16}$ .
Apply the fourth root.
$\left(\tfrac{81}{16}\right)^{1/4} = \frac{81^{1/4}}{16^{1/4}} = \frac{3}{2} = 1.5$
because $3^4 = 81$ and $2^4 = 16$ . Trap: dividing by 4 gives $\frac{81}{64} \approx 1.27$ .

Symbolic: given $7^n = x$ , find $7^{n/2}$ .

Apply the rule. $7^{n/2} = (7^n)^{1/2}$ .
Substitute. Since $7^n = x$ :
$7^{n/2} = x^{1/2} = \sqrt{x}$
Trap: $\tfrac{x}{2}$ , which confuses the index 1/2 with multiplying by one half.

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Common questions

Why does $9^{1/2}$ equal 3 and not 4.5?: Because the index 1/2 means square root, not divide by 2. The rule $a^{1/2} = \sqrt{a}$ says: take the square root. $\sqrt{9} = 3$ because $3^2 = 9$ . The answer 4.5 = 9/2 divides by 2 instead, which is the wrong operation.
What does the denominator of a fractional index tell you?: The denominator tells you which root to take: $a^{1/n} = \sqrt[n]{a}$ . So $a^{1/2}$ is the square root, $a^{1/3}$ is the cube root, and $a^{1/4}$ is the fourth root. The denominator is never a divisor.
Given $7^n = x$ , why is $7^{n/2}$ equal to $\sqrt{x}$ and not $\tfrac{x}{2}$ ?: Because $7^{n/2} = (7^n)^{1/2}$ (index 1/2 means square root), and substituting $7^n = x$ gives $7^{n/2} = x^{1/2} = \sqrt{x}$ . Writing $x/2$ would mean halving x, which is not what $a^{1/2} = \sqrt{a}$ says.

Related misconceptions

Negative indices mean reciprocals, not negative numbersReading the minus in 8^-5 as a sign on the result instead of recognising a^-n = 1/a^n, a companion index-laws misconception in the same Higher vertical.
Index laws operate on the index, not the baseApplying an index law by multiplying or changing the base instead of operating on the indices, the third misconception in the Higher index-laws vertical.

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