Operations on real numbers and the laws of exponents

Here (8=2^3) and (4^2=(2^2)^2=2^4), so (\dfrac{2^5 \times 2^3}{2^4}=2^4=16). In exams, converting numbers to the same base is useful.

By exponent laws, ((2^3)^2=2^6) and (2^6 \times 2^{-4}=2^2=4). In exams, add exponents when the base is the same.

Here (5^0=1), (3^{-1}=\dfrac{1}{3}), and (2^{-2}=\dfrac{1}{4}), so the value is (\dfrac{16}{3}). In exams, first convert negative exponents into fractions.

The numerator gives (a^m \times a^{2m}=a^{3m}), and then (\dfrac{a^{3m}}{a^{3m-2}}=a^2). In exams, subtract exponents during division.

The outside power (-2) multiplies both exponents, so (x^4y^{-6}=\dfrac{x^4}{y^6}). In exams, apply the outside power to every factor inside the bracket.

Since (9^{\frac{3}{2}}=(\sqrt{9})^3=3^3=27). In exams, connect the exponent (\dfrac{1}{2}) with square root.

Here (16^{\frac{1}{4}}=2), so (16^{\frac{3}{4}}=8) and (16^{-\frac{3}{4}}=\dfrac{1}{8}). In exams, a negative exponent means reciprocal.

Because (\sqrt{50}=5\sqrt{2}), (\sqrt{8}=2\sqrt{2}), and (\sqrt{18}=3\sqrt{2}), the answer is (4\sqrt{2}). In exams, combine only like surd terms.

Multiplying by (\sqrt{3}+\sqrt{2}) makes the denominator (3-2=1). In exams, remember to multiply by the conjugate.

\(\left(\dfrac{2}{3}\right)^{-2}=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\), so the product is (1). In exams, a fraction is inverted under a negative exponent.

Here (27^{\frac{2}{3}}=9) and (81^{\frac{1}{4}}=3), so the product is (27). In exams, first take the root and then apply the power.

The governing concept is equality of powers with the same positive base. Express 32 as a power of 2: \(32=2^5\). The equation therefore becomes \(2^{x+1}=2^5\). Since the base 2 is the same on both sides and is not 0 or 1, the exponents must be equal, giving \(x+1=5\). Subtracting 1 from both sides yields \(x=4\). Substitution confirms the result: \(2^{4+1}=2^5=32\). Thus option A is correct. Option B gives the exponent \(x+1\), not \(x\); option C results from subtracting incorrectly, and option D adds rather than subtracts 1.

From (9=3^2), (a=2), and from (8=2^3), (b=3), so (a+b=5). In exams, remembering small powers gives faster solutions.

Inside, (a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2), so (\dfrac{(a^2)^2}{a^3}=a). In exams, solve fractional exponents using the usual exponent rules.

Because (x^2-y^2=(x-y)(x+y)), the simplified form is (x+y). In exams, identifying difference of squares is very useful.

On expansion, ((p+q)^2=p^2+2pq+q^2) and ((p-q)^2=p^2-2pq+q^2), so the difference is (4pq). In exams, apply standard identities directly.

The product of coefficients (2) and (-3) is (-6), and powers of like variables are added. In exams, watch both the sign and the exponents carefully.

The coefficient is (\dfrac{6}{2}=3), (a^{3-1}=a^2), and (b^{2-(-1)}=b^3). In exams, the sign changes when subtracting a negative exponent.

The numerator is ((x^3)^2=x^6) and the denominator is (x^{-1}x^4=x^3), so the answer is (x^3). In exams, apply an exponent law at each step.

Since (0.00032=3.2\times 10^{-4}), (\dfrac{3.2\times 10^{-4}}{10^{-5}}=3.2\times 10^1=32). In exams, converting decimals to scientific notation helps.