Exponents Grade VIII Worksheet with solutions

Solutions

1. Question: By what number should \( \left(\frac{2}{7}\right)^{-2} \) be divided so that the quotient becomes 49?

   Solution: Let's call the number by which we should divide \(x\). Then the equation becomes \(\frac{\left(\frac{2}{7}\right)^{-2}}{x} = 49\). Simplifying, we get \(x = \frac{1}{49}\).

2. Question: Find x so that \( \left(\frac{3}{8}\right)^{14} \times \left(\frac{8}{3}\right)^{-9} = \left(\frac{3}{8}\right)^{2x-1} \).

   Solution: We can equate the powers because the bases are equal. This gives us \(14 - 9 = 2x - 1\). Solving for \(x\), we find \(x = 12\).

3. Question: Find the value of a/b if \( \left(\frac{a}{b}\right)^{-6} = \left(\frac{2}{7}\right)^{-6} \times \left(\frac{14}{9}\right)^{-6} \).

   Solution: Since the left-hand side and the right-hand side are equal, we can equate the bases which gives us \( \frac{a}{b} = \frac{2}{7} \times \frac{14}{9} = \frac{4}{3} \). So, \( a = 4 \) and \( b = 3 \).

4. Question: Simplify \( \frac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}} \).

   Solution: This simplifies to \( \frac{125 \times 5^2 \times 6^5}{3^5} = \frac{3125 \times 7776}{243} = 100000 \).

5. Question: The mass of Earth is \( 5.97 \times 10^{24} \) kg and the mass of the Moon is \( 7.35 \times 10^{22} \) kg. What is the total mass?

   Solution: The total mass is the sum of the mass of the Earth and the Moon, which is \( 5.97 \times 10^{24} + 7.35 \times 10^{22} = 5.97 \times 10^{24} + 0.0735 \times 10^{24} = 6.0435 \times 10^{24} \) kg.