Operations on real numbers and the laws of exponents

Here (169^{-1}=13^{-2}) and (2197^{-1}=13^{-3}), so (\frac{13^{4}\cdot13^{-2}}{13^{-3}}=13^{5}). In exams, division by a negative power adds the exponent.

We have (\sqrt{242}=11\sqrt{2}), (\sqrt{128}=8\sqrt{2}), (\sqrt{98}=7\sqrt{2}), and (\sqrt{72}=6\sqrt{2}). The total is (4\sqrt{2}).

The governing concept is conversion to a common base and then comparison of exponents. Since 27 = 3^3, we have 27^(x−1) = (3^3)^(x−1) = 3^(3x−3). Therefore the left-hand side becomes 3^x · 3^(3x−3) = 3^(4x−3). Also, 243 = 3^5. The bases are equal and the base 3 is positive and different from 1, so their exponents must be equal: 4x − 3 = 5. Adding 3 gives 4x = 8, and dividing by 4 gives x = 2. Substitution confirms the result: 3^2 · 27^1 = 9 · 27 = 243. Hence option B is correct. Option A leaves the exponent too small, while C and D do not satisfy the original equation.

Let (A=x^{-2}) and (B=y^{-2}). Then (\frac{A^{2}-B^{2}}{A-B}=A+B), so the answer is (x^{-2}+y^{-2}=\frac{x^{2}+y^{2}}{x^{2}y^{2}}).

Because ((3+\sqrt{10})^{2}=9+10+6\sqrt{10}=19+6\sqrt{10}), (\sqrt{A}=3+\sqrt{10}). In exams, identify perfect-square surd forms.

Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).

Factoring (4^{x}), we get (4^{x}(1+4+16)=336). Thus (21\cdot4^{x}=336), (4^{x}=16), and (x=2).

Here \(25^{\frac{3}{2}}=(5^{2})^{\frac{3}{2}}=5^{3}\) and \(125^{-\frac{2}{3}}=(5^{3})^{-\frac{2}{3}}=5^{-2}\). The product is (5).

Using the square of a binomial, \(r^2=(\sqrt{21}+\sqrt{14})^2=21+14+2\sqrt{294}\). Since \(\sqrt{294}=\sqrt{49\times6}=7\sqrt{6}\), we get \(r^2=35+14\sqrt{6}\). Therefore, \(r^2-14\sqrt{6}=35\), so option C is correct. Exam tip: In such problems, simplify the cross-term \(2\sqrt{ab}\) carefully before subtracting.

We use (x^{10}-1024=(x^{5})^{2}-32^{2}=(x^{5}-32)(x^{5}+32)). Cancelling the common factor leaves (x^{5}+32).

The product of denominators is (26-25=1), and the numerator is ((\sqrt{26}+5)+(\sqrt{26}-5)=2\sqrt{26}). In exams, add conjugate fractions together.

Using the laws of exponents, \(x^{15}=(x^5)^3=3^3=27\) and \(x^{10}=(x^5)^2=3^2=9\). Hence, \(x^{15}+x^{10}=27+9=36\), so option B is correct. Exam tip: rewrite each exponent as a multiple of 5 before substituting the given value of \(x^5\).

Since \(\left(\frac{25}{49}\right)^{\frac{1}{2}}=\frac{5}{7}\), \(\left(\frac{25}{49}\right)^{-\frac{3}{2}}=\left(\frac{5}{7}\right)^{-3}=\frac{343}{125}\). In exams, take the square root first.

Here ((4\sqrt{3})^{2}=48), ((3\sqrt{5})^{2}=45), and the middle term is (24\sqrt{15}). Therefore, the expansion is (93-24\sqrt{15}).

We get (a=4), and (9^{b}=3^{2b}=3^{6}) gives (b=3). Thus (a^{b}-b^{a}=4^{3}-3^{4}=64-81=-17), which is not among the options.

Using the laws of exponents, \((5x^{-2})^2=25x^{-4}\) and \((2x^4)^2=4x^8\). Thus, the numerator becomes \(25x^{-4}\cdot4x^8=100x^4\). Therefore, \(\frac{100x^4}{20x^4}=5\), since \(x\ne0\). Hence, the correct answer is 5. Exam tip: square each bracket first, combine powers with the same base, and then cancel common factors.

The governing concept is expressing every quantity as a power of the same base, 10. We have 100 = 10^2, so 100^3 = (10^2)^3 = 10^6. Similarly, 1000 = 10^3, so 1000^2 = (10^3)^2 = 10^6. Substituting these values gives (10^k · 10^6)/10^6 = 10^k, because equal powers in the numerator and denominator cancel. The equation therefore reduces to 10^k = 10^5. Since the base 10 is positive and not equal to 1, equal powers have equal exponents, so k = 5. Thus option C is correct. The values 3, 4, and 6 would produce 10^3, 10^4, and 10^6 respectively after simplification, so none of them can equal 10^5.

Here (\sqrt{300}=10\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{108}=6\sqrt{3}). The numerator is (12\sqrt{3}), so the value is (12).

Since \((7+4\sqrt{3})(7-4\sqrt{3})=49-48=1\), we get \(\frac{1}{7+4\sqrt{3}}=7-4\sqrt{3}\). Therefore, \(y+\frac{1}{y}=(7+4\sqrt{3})+(7-4\sqrt{3})=14\). Exam tip: For expressions of the form \(a+b\sqrt{c}\), use the conjugate \(a-b\sqrt{c}\) to rationalise the denominator.

Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).