Operations on real numbers and the laws of exponents

The numerator is (a^{-1}+b^{-1}=\dfrac{a+b}{ab}) and the denominator is ((ab)^{-1}=\dfrac{1}{ab}), so the answer is (a+b). In exams, make a common denominator.

((2^5)^{\frac{2}{5}}=2^2=4) and ((3^3)^{\frac{1}{3}}=3), so the product is (12). In exams, apply the power of a power law.

The governing law is the power-of-a-product rule: when a product is raised to a common exponent n, the exponent applies separately to each factor. Thus (uv)^n=u^n v^n. For example, with u=2, v=3, and n=2, the left side is (2·3)^2=6^2=36, while the right side is 2^2·3^2=4·9=36. Therefore option A is correct. Option B wrongly changes multiplication into addition. Option C reverses the product rule and is not generally true; for instance, 2^2·3^2=36 but (2+3)^2=25. Option D combines different exponents without a valid exponent law; multiplication of powers with the same base, not different bases, is where exponents are added.

Here \(\left(2x^{-3}\right)^{-2}=2^{-2}x^{6}=\frac{x^{6}}{4}\), so multiplying by \(x^{-1}\) gives \(\frac{x^{5}}{4}\). In exams, first convert negative exponents carefully.

The numerator exponent is ((m+2)+(3-m)=5), and (\frac{a^{5}}{a^{4}}=a). In exams, add and subtract exponents only for the same base.

We have \(\left(3^{x}\right)^{2}=3^{2x}\) and \(729=3^{6}\), so \(2x=6\) and \(x=3\). In exams, rewrite both sides with the same base.

Since (25^{-2}=5^{-4}) and (125=5^{3}), (\frac{5^{8}\cdot5^{-4}}{5^{3}}=5^{1}=5). In exams, convert (25) and (125) into powers of (5).

Here (pq=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=3-2=1). In exams, use ((a+b)(a-b)=a^{2}-b^{2}).

To rationalize, (\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}). In exams, multiply by the conjugate.

Here (x=\frac{8}{9}), so (x^{-1}=\frac{9}{8}). In exams, apply (a^{-n}=\frac{1}{a^{n}}) in the correct direction.

Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.

Here (2^{x+1}+2^{x}=2\cdot2^{x}+2^{x}=3\cdot2^{x}=48), so (2^{x}=16=2^{4}). In exams, factor the common power (2^{x}).

We get (\sqrt{50}=5\sqrt{2}), (\sqrt{18}=3\sqrt{2}), and (\sqrt{8}=2\sqrt{2}), so the result is (6\sqrt{2}). In exams, combine only like surd terms.

The governing concept is the quotient law of exponents: for a non-zero common base, x^a/x^b=x^(a−b). Comparing the first condition with x^7 gives a−b=7. The second condition supplies a+b=13. Adding these two linear equations eliminates b: (a−b)+(a+b)=7+13, so 2a=20 and a=10. A check gives b=3; then a−b=10−3=7 and a+b=10+3=13, confirming both original conditions. Therefore option C is correct. Option A is actually the resulting value of b, option B is the exponent difference rather than a, and option D is the sum a+b, not the requested individual exponent.

Since \(\left(\frac{27}{8}\right)^{\frac{1}{3}}=\frac{3}{2}\), \(\left(\frac{27}{8}\right)^{-\frac{2}{3}}=\left(\frac{3}{2}\right)^{-2}=\frac{4}{9}\). In exams, take the cube root first.

Rationalizing gives (\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2). In exams, use the conjugate of the denominator.

Here \(\left(9^{2}\right)^{3}=(3^{2})^{6}=3^{12}\), and \(3^{12}\div3^{10}=3^{2}\). In exams, write (9) as \(3^{2}\).

We have \(\left(ab^{-2}\right)^{3}=a^{3}b^{-6}\), and multiplying by \(a^{-1}b^{5}\) gives \(a^{2}b^{-1}\). In exams, add exponents separately for each variable.

Using (x^{2}-y^{2}=(x-y)(x+y)), we get (x-y=2\sqrt{2}) and (x+y=6), so the value is (12\sqrt{2}). In exams, identities reduce calculation.

Here ((2^{5})^{3}=2^{15}), (4^{-2}=2^{-4}), and (8^{2}=2^{6}), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).