Exponent & Root: Level 2

Solution (C): Factor out the smallest term, \(3^x\). \(3^x(1 + 3^1) = 108\) \(3^x(1 + 3) = 108\) \(3^x(4) = 108\) \(3^x = 108 / 4 = 27\) Since \(27 = 3^3\), \(x = 3\).

Solution (C): Convert all terms to base 2. \(8 = 2^3\) and \(4 = 2^2\). \((2^3)^x \times 2^2 = 2^{11}\) \(2^{3x} \times 2^2 = 2^{11}\) \(2^{(3x + 2)} = 2^{11}\) Equate the exponents: \(3x + 2 = 11 \implies 3x = 9 \implies x = 3\).

Solution (B): Simplify the radicals by finding perfect square factors. 1. \( \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \). 2. \( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \). 3. Add them: \( 6\sqrt{2} + 5\sqrt{2} = 11\sqrt{2} \).

Solution (C): The denominator (3) is the root, and the numerator (2) is the power. \( (\sqrt[3]{27})^2 \). First, \( \sqrt[3]{27} = 3 \). Then, \( (3)^2 = 9 \).

Solution (C): 1. Solve for \(x\): \(2^{2x} = 16\). Since \(16 = 2^4\), we have \(2^{2x} = 2^4\). 2. \(2x = 4 \implies x = 2\). 3. Now find the value of \(2^{x+1}\): \(2^{(2+1)} = 2^3 = 8\).

Solution (C): Factor out \(5^x\) from the numerator. \( \frac{5^x(5^2 – 1)}{5^x} \). Cancel the \(5^x\) terms: \( 5^2 – 1 = 25 – 1 = 24 \).

Solution (C): Find the prime factorization of 72. \(72 = 8 \times 9 = 2^3 \times 3^2\). So, \(2^x \times 3^y = 2^3 \times 3^2\). By comparing the exponents, \(x=3\) and \(y=2\). \(x + y = 3 + 2 = 5\).

Solution (D): 1. Evaluate the numerator: \( \sqrt{100} = 10 \). 2. Evaluate the denominator: \( 10^{-1} = \frac{1}{10} \). 3. Divide: \( \frac{10}{\frac{1}{10}} = 10 \times \frac{10}{1} = 100 \).

Solution (C): Convert both sides to base 2. \(4 = 2^2\) and \(8 = 2^3\). \((2^2)^x = 2^3\) \(2^{2x} = 2^3\) Equate the exponents: \(2x = 3 \implies x = \frac{3}{2} = 1.5\).

Solution (C): Factor out the smallest term, \(5^{x-1}\). \(5^{x-1}(5^1 – 1) = 100\) \(5^{x-1}(5 – 1) = 100\) \(5^{x-1}(4) = 100\) \(5^{x-1} = 100 / 4 = 25\) Since \(25 = 5^2\), we have: \(5^{x-1} = 5^2\) \(x-1 = 2 \implies x = 3\).