Maths exponents worksheets grade 8 cbse - Free Printable

Here are the step-by-step solutions for the exponent problems on the worksheet.

1) $(5^2 \cdot 5^6) \div (5^4)^2$
* First, simplify the part inside the parentheses using the product rule ($a^m \cdot a^n = a^{m+n}$): $5^2 \cdot 5^6 = 5^{2+6} = 5^8$.
* Next, simplify the denominator using the power of a power rule ($(a^m)^n = a^{m \cdot n}$): $(5^4)^2 = 5^{4 \cdot 2} = 5^8$.
* Now divide: $5^8 \div 5^8 = 1$.

2) $(-2)^5 \cdot (-2)^3 \cdot (-2)^2 \cdot (-2)^0$
* Use the product rule to add all the exponents together: $5 + 3 + 2 + 0 = 10$.
* The expression becomes $(-2)^{10}$.
* Since the exponent (10) is an even number, the negative sign disappears.
* $2^{10} = 1024$.

3) $(3^2 \cdot 3)^2 + [(-2)^5 \div (-2)^3]^3$
* Simplify the first part: Inside the bracket is $3^2 \cdot 3^1 = 3^3$. Then square it: $(3^3)^2 = 3^6$.
* $3^6 = 729$.
* Simplify the second part: Inside the bracket is $(-2)^5 \div (-2)^3 = (-2)^{5-3} = (-2)^2$.
* $(-2)^2 = 4$.
* Cube that result: $4^3 = 64$.
* Add them together: $729 + 64 = 793$.

4) $(-9)^7 \div (-9)^6$
* Use the quotient rule ($a^m \div a^n = a^{m-n}$): $7 - 6 = 1$.
* The result is $(-9)^1$, which is simply $-9$.

5) $\frac{(3^2 \cdot 3^{-1})^3}{3}$
* Simplify inside the parentheses first: $3^2 \cdot 3^{-1} = 3^{2-1} = 3^1 = 3$.
* Cube that result: $3^3 = 27$.
* Divide by the denominator (which is $3^1$): $27 \div 3 = 9$.

6) $\frac{(3^{-2})^3}{4^2}$
* Simplify the numerator: $(3^{-2})^3 = 3^{-6} = \frac{1}{3^6}$.
* Simplify the denominator: $4^2 = 16$.
* Combine them: $\frac{1/3^6}{16} = \frac{1}{3^6 \cdot 16}$.
* Calculate the values: $3^6 = 729$.
* $729 \cdot 16 = 11,664$.
* The fraction is $\frac{1}{11664}$.

7) $\left(\frac{3^2 \cdot 3^{-1}}{2^2 \cdot 3^{-2}}\right)^{-1}$
* Simplify the top (numerator): $3^2 \cdot 3^{-1} = 3^1 = 3$.
* Simplify the bottom (denominator): $2^2 \cdot 3^{-2} = 4 \cdot \frac{1}{9} = \frac{4}{9}$.
* Divide the top by the bottom: $3 \div \frac{4}{9} = 3 \cdot \frac{9}{4} = \frac{27}{4}$.
* Apply the outer exponent of $-1$, which means taking the reciprocal (flipping the fraction): $\frac{4}{27}$.

8) $\frac{(-3)^2 \cdot (-2)^2}{7^2} \cdot 2^3$
* Calculate the powers in the numerator: $(-3)^2 = 9$ and $(-2)^2 = 4$.
* Multiply them: $9 \cdot 4 = 36$.
* Calculate the denominator: $7^2 = 49$.
* So the fraction is $\frac{36}{49}$.
* Multiply by the last term $2^3 = 8$: $\frac{36}{49} \cdot 8$.
* $36 \cdot 8 = 288$.
* The result is $\frac{288}{49}$.

9) $(-3)^3 \div 3 \cdot 2^4$
* Follow the order of operations (left to right for multiplication/division).
* First, $(-3)^3 = -27$.
* Divide by 3: $-27 \div 3 = -9$.
* Calculate $2^4 = 16$.
* Multiply: $-9 \cdot 16 = -144$.

10) $(-7)^0 + 8 \cdot (-2)^3$
* Anything to the power of 0 is 1, so $(-7)^0 = 1$.
* Calculate the power: $(-2)^3 = -8$.
* Multiply: $8 \cdot (-8) = -64$.
* Add the first part: $1 + (-64) = -63$.

Final Answer:
1) 1
2) 1024
3) 793
4) -9
5) 9
6) 1/11664
7) 4/27
8) 288/49
9) -144
10) -63