SAT Exponents Practice with Answers + Explanations

Rewrite as \(\left(\frac{81}{16}\right)^{\frac{1}{4}}\). Since \(81 = 3^4\) and \(16 = 2^4\), \(\left(\frac{3^4}{2^4}\right)^{\frac{1}{4}} = \frac{3}{2}\).

Rewrite 16 as \(2^4\): \(2^{y+3} = 2^4 \times 2^2 = 2^6\). Equating exponents gives \(y + 3 = 6\), so \(y = 3\).

Simplify the expression inside: \(8^{\frac{2}{3}} = 4\) and \(4^{\frac{1}{2}} = 2\), so \(\frac{4}{2} = 2\). Raising to the third power gives \(2^3 = 8\).

Raise both sides to the reciprocal of \(\frac{3}{4}\), which is \(\frac{4}{3}\), to isolate \(a\): \(a = 16^{\frac{4}{3}} = \left(2^4\right)^{\frac{4}{3}} = 2^{\frac{16}{3}} = 32\sqrt[3]{2}\).

Simplify inside the parentheses: \(3^2 \times 3^{-1} = 3^{2-1} = 3^1\). Then divide by \(3^3\): \(\frac{3^1}{3^3} = 3^{1-3} = 3^{-2}\). Now square this result: \((3^{-2})^2 = 3^{-4}\).

Rewrite \(27\) as \(3^3\), so \(27^{-\frac{2}{3}} = (3^3)^{-\frac{2}{3}} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}\).

\( \frac{1}{25} = 25^{-1} = (5^2)^{-1} = 5^{-2}\). Since the bases are equal, we can set the exponents equal: \(2x = -2\). Solving for \(x\) gives \(x = -1\).

First, simplify the expression inside the parentheses: \(2^4 \times 2^{-2} = 2^{4 - 2} = 2^2\). Now, raise it to the power of 3: \((2^2)^3 = 2^{2 \times 3} = 2^6\). Finally, divide by \(2^5\): \(\frac{2^6}{2^5} = 2^{6 - 5} = 2^1 = 2\).

Rewrite 16 as \(2^4\): \(2^{y+3} = 2^4 \times 2^2 = 2^6\). Equating exponents gives \(y + 3 = 6\), so \(y = 3\).

Rewrite as \(\left(\frac{81}{16}\right)^{\frac{1}{4}}\). Since \(81 = 3^4\) and \(16 = 2^4\), \(\left(\frac{3^4}{2^4}\right)^{\frac{1}{4}} = \frac{3}{2}\).

Simplify inside the parentheses: \(16^{\frac{1}{2}} = 4\), \(2^3 = 8\), and \(8^{\frac{2}{3}} = 4\). This gives \(\frac{4 \times 8}{4} = 8\). Now, \(8^2 = 64\).

\(\frac{1}{9} = 3^{-2}\). So, \((3^{x-1})^2 = 3^{-2}\) implies \(x - 1 = -1\). Thus, \(x = 0\).

Simplify inside the parentheses: \(5^3 \times 5^{-2} = 5^{3-2} = 5^1\). Now, \((5^1)^4 = 5^4\). Dividing by \(5^5\) gives \(5^{4-5} = 5^{-1} = \frac{1}{5}\).

Combine the exponents: \(k^{-3+2+1} = k^0 = 16\). Since \(k^0 = 1\) for any \(k \neq 0\), there’s no solution where \(k = 16\).

Rewrite \(64 = 2^6\) and \(8 = 2^3\). This gives \((2^6)^{\frac{5}{6}} \div (2^3)^{\frac{2}{3}} = 2^5 \div 2^2 = 2^{5-2} = 2^3 = 8\).