Skip to content
Blogs
Grade 8 Exponents and Powers: 10 Challenging Questions to Test Your Skills

Grade 8 Exponents and Powers: 10 Challenging Questions to Test Your Skills

September 5, 2026

These 10 questions are designed to feel like the kind of problems you might encounter in challenging Grade 8 math tests and competitions. They aren’t about simply plugging numbers into a formula. Each question has a small twist, rewriting numbers with a common base, spotting a repeating pattern, or turning a tricky equation into something much simpler.

No calculator is needed. The goal is to think, spot the pattern, and choose the right strategy. Try each question on your own before checking the step-by-step solution


Question 1

Which is larger?

$$3^{4444} \quad \text{or} \quad 4^{3333}$$

Justify your answer without calculating either number.

Step by step:

  1. Both exponents share a common factor: $4444 = 4 \times 1111$ and $3333 = 3 \times 1111$.
  2. Rewrite each number using that shared factor: $3^{4444} = (3^4)^{1111} = 81^{1111}$, and $4^{3333} = (4^3)^{1111} = 64^{1111}$.
  3. Now both numbers are raised to the same power, $1111$ — so we only need to compare the bases: $81 > 64$.

Answer: $3^{4444}$ is larger.

💡 Key rule: To compare two huge powers, look for a shared factor in the exponents. Rewrite both numbers with that same outer exponent, then just compare the smaller bases.



Question 2

A computer processes $2^{40}$ operations per second. About how many operations does it process in:

$$2^{-10} \text{ seconds}$$

Step by step:

  1. Total operations = rate × time = $2^{40} \times 2^{-10}$.
  2. Same base, so multiplying means adding the exponents: $40 + (-10) = 30$.
  3. That gives $2^{30}$ operations.
  4. $2^{30} = 1{,}073{,}741{,}824$.

Answer: $2^{30} = 1{,}073{,}741{,}824$ operations.

💡 Key rule: A negative exponent on “time” just means a fraction of a second — you still add exponents the same way when multiplying.



Question 3

Find the units digit of:

$$7^{2026}$$

Step by step:

  1. List the units digits of powers of 7: $7^1=7$, $7^2=49$, $7^3=343$, $7^4=2401$ → the units digits are $7, 9, 3, 1$, and then the pattern repeats.
  2. The pattern has length 4, so divide the exponent by 4: $2026 \div 4 = 506$ remainder $2$.
  3. A remainder of $2$ means the units digit matches the 2nd number in the pattern: $9$.

Answer: $9$

💡 Key rule: When an exponent is too big to calculate directly, look for a repeating pattern in the units digit instead.



Question 4

If:

$$2^x = 3^y = 6^z$$

and $x$, $y$, $z$ are all nonzero, find the value of $\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{z}$.

Step by step:

  1. Let $2^x = 3^y = 6^z = k$, for some number $k$.
  2. That means $2 = k^{1/x}$, $3 = k^{1/y}$, and $6 = k^{1/z}$.
  3. Since $6 = 2 \times 3$, we can write: $k^{1/z} = k^{1/x} \times k^{1/y} = k^{1/x + 1/y}$.
  4. Same base $k$ on both sides, so the exponents must match: $\dfrac{1}{z} = \dfrac{1}{x} + \dfrac{1}{y}$.
  5. Rearranging gives $\dfrac{1}{x} + \dfrac{1}{y} - \dfrac{1}{z} = 0$.

Answer: $0$

💡 Key rule: When one number is the product of two others (like $6 = 2 \times 3$), their “exponent relationships” combine the same way.



Question 5

Simplify:

$$\left(x^{a-b}\right)^{a+b} \times \left(x^{b-c}\right)^{b+c} \times \left(x^{c-a}\right)^{c+a}$$

Step by step:

  1. Apply “power of a power” to the first term: $\left(x^{a-b}\right)^{a+b} = x^{(a-b)(a+b)} = x^{a^2-b^2}$.
  2. The same way, the second term becomes $x^{b^2-c^2}$, and the third becomes $x^{c^2-a^2}$.
  3. Multiply all three (same base, so add the exponents): $(a^2-b^2) + (b^2-c^2) + (c^2-a^2)$.
  4. Everything cancels out to $0$.
  5. So the whole expression is $x^0$.

Answer: $1$

💡 Key rule: $(a-b)(a+b) = a^2 - b^2$. Spotting this pattern turns a scary-looking expression into one where everything cancels.



Question 6

A bacteria population doubles every 3 hours, starting at 500. Write an expression for the population after $t$ hours, and find the population after 15 hours.

Step by step:

  1. “Doubles every 3 hours” means the population is multiplied by $2$ for every 3-hour block that passes.
  2. In $t$ hours, the number of 3-hour blocks is $t/3$.
  3. So the population is $500 \times 2^{t/3}$.
  4. After 15 hours: $t/3 = 15/3 = 5$.
  5. Population = $500 \times 2^5 = 500 \times 32$.

Answer: $500 \times 2^{t/3}$; after 15 hours, $16{,}000$ bacteria.

💡 Key rule: “Doubles every X hours” always becomes $2^{\,\text{time} \div X}$ in the exponent.



Question 7

If:

$$a^2 = b^3 = c^5$$

where $a$, $b$, $c$ are positive integers and all powers of the same prime number, find the smallest possible value of $a + b + c$ (other than the trivial case $a=b=c=1$).

Step by step:

  1. Since $a$, $b$, $c$ are all powers of the same prime $p$, write $a = p^i$, $b = p^j$, $c = p^k$.
  2. Then $a^2 = b^3 = c^5$ means $2i = 3j = 5k$. Call this shared value $M$.
  3. For $i$, $j$, $k$ to be whole numbers, $M$ must be divisible by $2$, $3$, and $5$ — so the smallest option is their LCM: $2 \times 3 \times 5 = 30$.
  4. That gives $i = 15$, $j = 10$, $k = 6$.
  5. Using the smallest possible prime, $p = 2$: $a = 2^{15} = 32{,}768$, $b = 2^{10} = 1{,}024$, $c = 2^{6} = 64$.
  6. Add them: $32{,}768 + 1{,}024 + 64$.

Answer: $33{,}856$

💡 Key rule: When several powers all need to be equal, think in exponents and find the LCM — it hands you the smallest whole-number solution.



Question 8

Find the integer $n$ for which:

$$2^n + 2^{n+1} + 2^{n+2} = 224$$

Step by step:

  1. Factor $2^n$ out of every term: $2^n(1 + 2 + 4) = 2^n \times 7$.
  2. So $2^n \times 7 = 224$.
  3. Divide both sides by 7: $2^n = 32$.
  4. $32 = 2^5$, so $n = 5$.

Answer: $n = 5$

💡 Key rule: When the same base shows up in a sum, factor it out first. Turning addition into multiplication makes the equation solvable.



Question 9

Simplify:

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

Step by step:

  1. $\left(-\frac{1}{2}\right)^{-3}$ means “flip the fraction, then cube it”: $(-2)^3 = -8$.
  2. Square that: $(-8)^2 = 64$.
  3. $\left(-\frac{1}{2}\right)^{2} = \frac{1}{4}$ (a negative number squared becomes positive).
  4. Raise $\frac{1}{4}$ to the power $-3$: flip it and cube it: $4^3 = 64$.
  5. Divide: $64 \div 64$.

Answer: $1$

💡 Key rule: Watch your signs carefully — an even outer exponent always makes the result positive, no matter what sign was inside.



Question 10

Arrange in increasing order, without a calculator:

$$2^{300}, \quad 3^{200}, \quad 5^{100}$$

Step by step:

  1. The exponents $300$, $200$, and $100$ all share a common factor of $100$.
  2. Rewrite each number using that shared exponent: $2^{300} = (2^3)^{100} = 8^{100}$, and $3^{200} = (3^2)^{100} = 9^{100}$.
  3. $5^{100}$ already has the exponent $100$.
  4. Now all three numbers share the same outer exponent, so just compare the bases: $5 < 8 < 9$.

Answer: $5^{100} < 2^{300} < 3^{200}$

💡 Key rule: When comparing powers with different bases and different exponents, rewrite them so they all share the same exponent — then it’s just a comparison of bases.



Quick Answer Key

QuestionAnswer
1$3^{4444}$ is larger
2$2^{30} = 1{,}073{,}741{,}824$
3$9$
4$0$
5$1$
6$500 \times 2^{t/3}$; $16{,}000$ after 15 hours
7$33{,}856$
8$n = 5$
9$1$
10$5^{100} < 2^{300} < 3^{200}$

Now you try: 5 multiple-choice questions

You’ve worked through the 10 exam-style problems. These 5 are designed to test your knowledge - pick an answer, then the next question appears.

Check your understanding

5 multiple-choice questions

Question 1 of 5


New to this topic? Start with the warm-up set first: Exponents and Powers Class 8: 5 Challenging Questions (With Simple Step-by-Step Solutions).