Hard GCSE Fractional indices Questions

Understand the goal

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

Write the square root as a power

A square root is the power one half.

Write the cube root as a power

A cube root is the power one third.

Rewrite the product

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

Recall the product law

Multiplying powers of the same base means adding the indices.

Set up the index addition

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

Find a common denominator

Sixths work for both fractions.

Add the fractions

Three sixths plus two sixths is five sixths.

Write the single power

So the product is x to the power five sixths.

Interpret the result

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

Check with a value: set $x = 64$

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

Multiply the check values

The product is $32$.

Evaluate the formula at $x = 64$

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

Watch for the common error

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

State the answer

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

Read the index carefully

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

Deal with the negative sign first

A negative index on a fraction flips it over.

Recall the fraction-base rule

Apply the power to top and bottom separately.

Split the fraction

Two fractional powers to evaluate.

Take the fourth root of $16$

$2 \times 2 \times 2 \times 2 = 16$, so the fourth root of $16$ is $2$.

Cube it

The numerator becomes $8$.

Take the fourth root of $81$

$3 \times 3 \times 3 \times 3 = 81$, so the fourth root of $81$ is $3$.

Cube it

The denominator becomes $27$.

Combine the parts

Put the evaluated top over the evaluated bottom.

Check the flip direction

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

Shortcut view

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

Check by inverting the answer

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

Watch for the common error

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

Reflect on the order used

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

State the answer

The value is eight twenty-sevenths.

Look at the structure of the equation

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

Evaluate the right side first: fifth root

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

Raise to the power $4$

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

Rewrite the equation

The equation is now much simpler.

Write $16$ as a power of $2$

Convert the remaining numbers to base $2$.

Substitute into the left side

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

Recall the power-of-a-power law

Multiply the indices in the bracket.

Simplify the bracket

4 times x gives 4x.

Recall the product law

Multiplying powers of the same base means adding the indices.

Combine the left side

Add the indices $4x$ and $1$.

Equate the indices

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

Subtract $1$ from both sides

Remove the constant term first.

Solve for x

Divide both sides by 4: x is three quarters.

Check the solution

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

State the answer

The solution is x equals three quarters.

Understand the goal

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

Bracket with square numbers

$10$ sits between the square numbers $9$ and $16$.

Take roots of the bracket

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

Try $3.1$

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

Try $3.2$

$3.2$ squared is $10.24$, which is past $10$.

Narrow the bracket

Root $10$ is trapped between $3.1$ and $3.2$.

Decide which end is closer

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

Test the midpoint region

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

Confirm with $3.17$

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

Round to one decimal place

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

State the estimate

So $10$ to the power one half is approximately $3.2$.

Sense check against a known root

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

Watch for the common error

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

Reflect on the method

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

State the answer

To one decimal place, the value is about $3.2$.

Understand the inequality

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.

Plan: remove the fraction

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

Square both sides

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

Evaluate $100$ squared

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

Recall a landmark power of $2$

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

Build up from the landmark

Keep doubling from $1024$.

Continue doubling

$8192$ is still below $10000$.

One more doubling

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

Bracket the answer

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

Translate back to the original

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

Evaluate the check

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

Check $n = 13$ fails

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

Estimate the root

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

Confirm minimality

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

State the answer

The smallest integer n is 14.