Operations on real numbers and the laws of exponents

The governing concept is the law of powers: (a^m)^n = a^(mn), together with rewriting a number in terms of the known base. Since 8 = 2^3, we have 8^x = (2^3)^x = (2^x)^3. The question gives 2^x = 3, so substitute 3 to obtain (2^x)^3 = 3^3 = 27. Therefore option A is correct. Option B, 9, would correspond to 3^2 and uses the wrong exponent; 24 is not produced by any valid power rule, and 6 incorrectly multiplies rather than raises the given value to the third power.

Because (a^{-2}=\dfrac{1}{a^2}=\dfrac{1}{5}), (\dfrac{1}{5}+5=\dfrac{26}{5}). In exams, write (a^{-2}) as (\dfrac{1}{a^2}).

(x^4-16=(x^2-4)(x^2+4)), so the simplified form is (x^2+4). In exams, treat (x^4) as ((x^2)^2) for factorisation.

When both expansions are added, (2mn) and (-2mn) cancel, giving (2m^2+2n^2). In exams, notice opposite middle terms.

Adding like terms gives (3x^2+x^2=4x^2), (-2x+5x=3x), and (1-4=-3). In exams, add like terms in columns.

Changing the signs of the second bracket gives (5x^3-2x+7-2x^3-3x+5), so the answer is (3x^3-5x+12). In exams, change the sign of every term in the bracket during subtraction.

By distribution, (6x^2+8x-3x-4=6x^2+5x-4). In exams, combine cross terms with the correct sign.

(x^{5-(-1)}=x^6) and (y^{-2-3}=y^{-5}), so the form is (\dfrac{x^6}{y^5}). In exams, simplify the exponent of each variable separately.

The expression inside is \(a^{-3}b^4\), and the power (-1) gives its reciprocal \(\dfrac{a^3}{b^4}\). In exams, apply the outer negative power at the end.

Since (\sqrt[3]{64}=4), (4^{-2}=\dfrac{1}{16}). In exams, first evaluate the root and then apply the negative exponent.

(2^{-1}+3^{-1}=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}), so the whole value is (\dfrac{6}{5}). In exams, simplify the denominator first.

Here (9^{-1}=3^{-2}) and (27^{-1}=3^{-3}), so the value is (3^{4-2-(-3)}=3^5=243). In exams, convert all terms to the same base.

The numerator is ((a^{-2}b^3)^2=a^{-4}b^6) and the denominator is ((ab^{-1})^{-1}=a^{-1}b), so the answer is (\dfrac{b^5}{a^3}). In exams, apply the outside power first.

Since (25^{\frac{3}{2}}=125) and (125^{\frac{2}{3}}=25), the value is (5). In exams, understand the root first in fractional powers.

Because (x^3-27=(x-3)(x^2+3x+9)), the quotient is (x^2+3x+9). In exams, identify the difference of cubes.

(\sqrt{75}=5\sqrt{3}), (\sqrt{12}=2\sqrt{3}), and (\sqrt{48}=4\sqrt{3}), so the answer is (7\sqrt{3}). In exams, combine only terms with the same radical part.

Multiplying by (\sqrt{7}-\sqrt{5}) makes the denominator (7-5=2) and gives (\sqrt{7}-\sqrt{5}). In exams, use the conjugate.

Since (4^{x+1}=2^{2x+2}) and (128=2^7), we get (2x+2=7) and (x=\dfrac{5}{2}). In exams, write both sides with the same base.

Rewrite \(16\) as \(2^4\): \(16^p=(2^4)^p=(2^p)^4=5^4=625\). Therefore, the correct answer is 625. The distractor 125 equals \(5^3\), but the required exponent is 4. Exam tip: Express the new base as a power of the given base and apply \((a^m)^n=a^{mn}\).

Because (x^2-y^2=(x-y)(x+y)), ((x+y)) cancels and (x-y) remains. In exams, identify difference of squares quickly.