Operations on real numbers and the laws of exponents

Correct answer: A. (2^{2}\cdot3)

Explanation: Since (6^{5}=2^{5}\cdot3^{5}), (\frac{2^{5}3^{5}}{2^{3}3^{4}}=2^{2}\cdot3). In exams, split a composite base into prime bases.

Correct answer: C. (4)

Explanation: The left side has exponents (2n) and (3n), so \(2n=8\) and \(3n=12\), giving \(n=4\). In exams, match exponents of both variables.

Correct answer: A. (4x^{2})

Explanation: Since (\sqrt[3]{64}=4) and (\sqrt[3]{x^{6}}=x^{2}), the answer is (4x^{2}). In exams, divide the exponent by (3) for cube roots.

Correct answer: A. \(13-2\sqrt{22}\)

Explanation: \((\sqrt{11}-\sqrt{2})^{2}=11+2-2\sqrt{22}=13-2\sqrt{22}\). In exams, include both \(+b^{2}\) and (-2ab) in \((a-b)^{2}\).

Correct answer: C. (81)

Explanation: From (2^{a}=2^{3}), (a=3), and from (3^{b}=3^{4}), (b=4), so (a^{b}=3^{4}=81). In exams, compare powers using equal bases.

Correct answer: A. (x^{2}-1)

Explanation: (\frac{x^{5}-x^{3}}{x^{3}}=\frac{x^{3}(x^{2}-1)}{x^{3}}=x^{2}-1). In exams, take out the common factor first.

Correct answer: A. \(\frac{1}{2}\)

Explanation: Here \(16^{-\frac{3}{4}}=(2^{4})^{-\frac{3}{4}}=2^{-3}\) and \(8^{\frac{2}{3}}=(2^{3})^{\frac{2}{3}}=2^{2}\), so the value is \(2^{-1}=\frac{1}{2}\). In exams, multiply powers of powers.

Correct answer: A. (2+\sqrt{3})

Explanation: Since ((2+\sqrt{3})^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), (\sqrt{A}=2+\sqrt{3}). In exams, recognize the form ((a+b)^{2}).

Correct answer: A. (x+y)

Explanation: Here (x^{-1}+y^{-1}=\frac{x+y}{xy}) and ((xy)^{-1}=\frac{1}{xy}), so division gives (x+y). In exams, converting negative exponents to fractions is safer.

Correct answer: A. (1)

Explanation: ((\sqrt{13}+\sqrt{12})(\sqrt{13}-\sqrt{12})=13-12=1). In exams, the product of conjugate surds is rational.

Correct answer: A. \(x^{-6}y^{5}\)

Explanation: \(\frac{3x^{-2}}{y^{-1}}=3x^{-2}y\), its cube is \(27x^{-6}y^{3}\), and multiplying by \(\frac{y^{2}}{27}\) gives \(x^{-6}y^{5}\). In exams, turn division by a negative power into multiplication.

Correct answer: B. (5)

Explanation: Since (4^{x}=(2^{2})^{x}=2^{2x}), (2x=10) and (x=5). In exams, convert mixed bases into a common base.

Correct answer: A. (1)

Explanation: (\sqrt{45}=3\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (\sqrt{5}), and division gives (1). In exams, first make like radicals.

Correct answer: A. \((\sqrt{3}-\sqrt{2})^{2}\)

Explanation: ((\sqrt{3}-\sqrt{2})^{2}=3+2-2\sqrt{6}=5-2\sqrt{6}). In exams, identify (a,b) from (a+b) and (2\sqrt{ab}).

Correct answer: A. (6)

Explanation: \(\left(\frac{2}{3}\right)^{-3}=\left(\frac{3}{2}\right)^{3}=\frac{27}{8}\) and \(\left(\frac{9}{4}\right)^{-1}=\frac{4}{9}\), so the product is (6). In exams, invert the fraction for negative powers.

Correct answer: A. (18)

Explanation: Here (x^{6}=(x^{2})^{3}=27) and (x^{4}=(x^{2})^{2}=9), so the difference is (18). In exams, express powers using the given (x^{2}).

Correct answer: A. (a^{-2}b)

Explanation: ((a^{2}b^{-1})^{-3}=a^{-6}b^{3}), then (\frac{a^{-6}b^{3}}{a^{-4}b^{2}}=a^{-2}b). In exams, subtract powers of the same base during division.

Correct answer: A. \(9x^{2}\)

Explanation: \(\left(81x^{4}\right)^{\frac{1}{2}}=\sqrt{81x^{4}}=9x^{2}\). In exams, the exponent becomes half under a square root.

Correct answer: B. (3)

Explanation: The total exponent on the left is (x+x+2-3=2x-1), and (32=2^{5}), so (2x-1=5), giving (x=3). In exams, convert the whole expression into one power.

Correct answer: A. (2\sqrt{3})

Explanation: The first term becomes (\sqrt{3}-\sqrt{2}), and the second becomes (\sqrt{3}+\sqrt{2}), so the sum is (2\sqrt{3}). In exams, rationalize both denominators separately.