Exponents and Powers: 5 Challenging Grade 8 Questions

These 5 questions are a step up from basic practice. Each one uses more than one exponent rule at a time, which is exactly how they tend to show up on real tests. Below, each question is followed by a simple, step-by-step solution, plus a quick callout with the one rule that makes it click.

Question 1

Simplify:

$$\frac{(2^{-3})^2 \times 2^7}{2^{-4} \times 2^2}$$

Step by step:

1. Start with $(2^{-3})^2$. Raising a power to another power means you multiply the exponents: $-3 \times 2 = -6$, so this becomes $2^{-6}$.
2. Multiply that by $2^7$. When bases are the same, multiplying means adding exponents: $-6 + 7 = 1$, giving $2^1$.
3. Now simplify the bottom: $2^{-4} \times 2^2 = 2^{-4+2} = 2^{-2}$.
4. Divide: $2^1 \div 2^{-2}$. Dividing means subtracting exponents: $1 - (-2) = 3$.

Answer: $2^3 = 8$

💡 Key rule: Multiplying same-base powers → add the exponents. Dividing → subtract them.

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Question 2

Evaluate:

$$\left(\frac{2}{3}\right)^{-4} \times \left(\frac{3}{2}\right)^{-2} \div \left(\frac{4}{9}\right)^{-1}$$

Step by step:

1. A negative exponent means “flip the fraction, then make the exponent positive.” So $\left(\frac{2}{3}\right)^{-4}$ becomes $\left(\frac{3}{2}\right)^{4} = \frac{81}{16}$.
2. The same way, $\left(\frac{3}{2}\right)^{-2}$ becomes $\left(\frac{2}{3}\right)^{2} = \frac{4}{9}$.
3. And $\left(\frac{4}{9}\right)^{-1}$ becomes $\frac{9}{4}$ — dividing by $\frac{9}{4}$ is the same as multiplying by $\frac{4}{9}$.
4. Multiply everything together: $\frac{81}{16} \times \frac{4}{9} \times \frac{4}{9} = \frac{9}{4} \times \frac{4}{9}$.

Answer: $1$

💡 Key rule: A negative exponent flips the fraction. It does not make the number negative.

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Question 3

Find the value of $x$:

$$5^{2x-1} \times 25 = 5^{4x-3} \div 5^{x-2}$$

Step by step:

1. Rewrite $25$ as a power of $5$: $25 = 5^2$.
2. Left side becomes $5^{2x-1} \times 5^{2} = 5^{2x+1}$ (add the exponents).
3. Right side becomes $5^{4x-3} \div 5^{x-2} = 5^{3x-1}$ (subtract the exponents).
4. Since both sides now have the same base, their exponents must be equal: $2x+1 = 3x-1$.
5. Solve: $x = 2$.

Answer: $x = 2$

💡 Key rule: If two powers with the same base are equal, their exponents must be equal too.

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Question 4

Simplify:

$$\frac{(a^{2b})^3 \times (a^{-b})^6}{(a^{3b})^2 \times a^0}$$

Step by step:

1. On top, apply “power of a power”: $(a^{2b})^3 = a^{6b}$ and $(a^{-b})^6 = a^{-6b}$.
2. Multiply those together: $a^{6b} \times a^{-6b} = a^{0} = 1$.
3. On the bottom, $(a^{3b})^2 = a^{6b}$, and remember — anything raised to the power $0$ equals $1$, so $a^0 = 1$.
4. Bottom becomes $a^{6b} \times 1 = a^{6b}$.
5. Divide: $1 \div a^{6b} = a^{-6b}$.

Answer: $a^{-6b}$

💡 Key rule: Any nonzero number or variable raised to the power $0$ equals $1$ — it does not make the whole expression $0$. That’s exactly the trap this question is testing.

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Question 5

Evaluate:

$$\left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right]^{-1} \times \left(\frac{2}{5}\right)^{0}$$

Step by step:

1. $\left(\frac{1}{3}\right)^{-2}$ means “flip and square”: $3^2 = 9$.
2. $\left(\frac{1}{2}\right)^{-3}$ means “flip and cube”: $2^3 = 8$.
3. Subtract: $9 - 8 = 1$.
4. Raise that to the power $-1$: flipping $1$ still gives $1$.
5. Multiply by $\left(\frac{2}{5}\right)^{0}$, which is just $1$.

Answer: $1$

💡 Key rule: Always simplify what’s inside the brackets first, before applying the exponent on the outside.

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Quick Answer Key

Question	Answer
1	8
2	1
3	x = 2
4	$a^{-6b}$
5	1

Want a harder set once these feel comfortable? Look out for Grade 8 Exponents and Powers: 10 Hard Questions to Test Your Skills