Exponent & Root: Level 1

Solution (D): When multiplying with the same base, we add the exponents. \(3^2 \times 3^3 = 3^{(2+3)} = 3^5 = 243\).

Solution (C): When dividing with the same base, we subtract the exponents. \( \frac{5^6}{5^4} = 5^{(6-4)} = 5^2 = 25 \).

Solution (D): When raising a power to another power, we multiply the exponents. \( (2^2)^3 = 2^{(2 \times 3)} = 2^6 = 64 \).

Solution (D): Any non-zero number to the power of 0 is 1. Any number to the power of 1 is itself. \( 7^0 + 7^1 = 1 + 7 = 8 \).

Solution (D): \( 4^{-1} = \frac{1}{4} \) and \( 4^0 = 1 \). \( \frac{1}{4} + 1 = \frac{1}{4} + \frac{4}{4} = \frac{5}{4} \).

Solution (D): We need to write 81 as a power of 3. \( 3^1 = 3 \) \( 3^2 = 9 \) \( 3^3 = 27 \) \( 3^4 = 81 \) So, \(3^x = 3^4\), which means \(x = 4\).

Solution (C): \( 2^x \times 2^1 = 16 \) \( 2^{(x+1)} = 16 \) Since \( 16 = 2^4 \), we have: \( 2^{(x+1)} = 2^4 \) \( x+1 = 4 \implies x = 3 \).

Solution (C): \( \sqrt{16} = 4 \). \( \sqrt{4} = 2 \). \( 4 \times 2 = 8 \).

Solution (B): To simplify \( \sqrt{50} \), find a perfect square factor. \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \).

Solution (B): \( (3^x)^2 = 3^{(x \times 2)} = 3^{2x} \). \( 81 = 3^4 \). So, \( 3^{2x} = 3^4 \). Equate the exponents: \( 2x = 4 \implies x = 2 \).