Operations on real numbers and the laws of exponents

Correct answer: A. (1)

Explanation: The direct answer is option A: 1. Evaluate each factor carefully. Since 64=4^3, \(64^{2/3}=(4^3)^{2/3}=4^2=16\). Since 8=2^3, \(8^{-4/3}=(2^3)^{-4/3}=2^{-4}=\frac{1}{16}\). Their product is \(16\cdot\frac{1}{16}=1\). Option A is therefore correct. Option B, 2, is wrong because the two factors cancel completely, not partially. Option C, 4, is wrong for the same reason: the product is not left with a factor of 4. Option D, \(\frac12\), is wrong because the negative exponent gives the reciprocal of 16, and that reciprocal is exactly cancelled by the first factor. The exponent rule \((a^m)^n=a^{mn}\) is used, and a negative exponent means \(a^{-r}=1/a^r\). A quick alternative is to express both numbers with base 2: 64=2^6, so the first factor is 2^4, while 8=2^3, so the second is 2^{-4}; their product is 2^0=1. Remember: negative exponents create reciprocals, and opposite exponents cancel.

Correct answer: C. 4

Explanation: Using the exponent rule \((ab)^k=a^k b^k\), the left-hand side becomes \(x^{-3k}y^{2k}\). Comparing the exponents of the same bases gives \(-3k=-12\) and \(2k=8\), and both equations yield \(k=4\). Exam tip: When monomials with the same bases are equal, compare the corresponding exponents.

Correct answer: A. (6a^{4}b^{3})

Explanation: We have (\sqrt[3]{216}=6), (\sqrt[3]{a^{12}}=a^{4}), and (\sqrt[3]{b^{9}}=b^{3}). In exams, divide exponents by (3) under a cube root.

Correct answer: A. (\frac{10}{3})

Explanation: Here (3^{-2}+3^{-4}=\frac{1}{9}+\frac{1}{81}=\frac{10}{81}), and (3^{-3}=\frac{1}{27}). Division gives (\frac{10}{3}).

Correct answer: C. 10

Explanation: Squaring the given expression, \(x^{2}=(\sqrt{7}-\sqrt{3})^{2}=7+3-2\sqrt{21}=10-2\sqrt{21}\). Therefore, \(x^{2}+2\sqrt{21}=10\), so option C is correct. Exam tip: While using \((a-b)^2=a^2+b^2-2ab\), take care to retain the negative sign in the middle term.

Correct answer: A. \(7r^{5}s^{-6}\)

Explanation: Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).

Correct answer: A. 0

Explanation: Use the conjugate-product identity \((a+b)(a-b)=a^2-b^2\). Thus, \((\sqrt{17}+\sqrt{8})(\sqrt{17}-\sqrt{8})=17-8=9\). Also, \(\sqrt{81}=9\), so the complete expression is \(9-9=0\). Exam tip: Identify conjugate pairs first and apply the difference-of-squares identity instead of expanding the radicals term by term.

Correct answer: C. 6

Explanation: Since \(18=2\cdot3^{2}\), we get \(18^{3}=2^{3}\cdot3^{6}\). Therefore, \(\frac{18^{3}}{2^{2}\cdot3^{5}}=\frac{2^{3}\cdot3^{6}}{2^{2}\cdot3^{5}}=2^{3-2}\cdot3^{6-5}=2\cdot3=6\). Hence, the correct answer is 6. Exam tip: when dividing powers with the same base, subtract the exponents using \(a^{m}\div a^{n}=a^{m-n}\).

Correct answer: A. (12\sqrt{10})

Explanation: Here (\frac{1}{s}=\sqrt{10}-3), so (s-\frac{1}{s}=6) and (s+\frac{1}{s}=2\sqrt{10}). Thus (s^{2}-\frac{1}{s^{2}}=12\sqrt{10}).

Correct answer: A. \(\frac{3x^{2}y^{3}}{4}\)

Explanation: We get \(\left(\frac{64x^{-6}}{27y^{9}}\right)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}\). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).

Correct answer: C. (8)

Explanation: We use \(x^{2}-\frac{1}{x^{2}}=\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\). Thus \(40=5\left(x+\frac{1}{x}\right)\), so the value is (8).

Correct answer: A. \(2b\)

Explanation: Combining like terms gives \(4b^{-2}+6b^{-2}=10b^{-2}\). Therefore, \(\frac{10b^{-2}}{5b^{-3}}=2b^{-2-(-3)}=2b\), so option A is correct. Exam tip: when dividing powers with the same nonzero base, subtract the exponents: \(b^m\div b^n=b^{m-n}\).

Correct answer: A. (192\sqrt{3})

Explanation: From (\sqrt{x}=4\sqrt{3}), (x=48), and (x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}). In exams, write (x^{\frac{3}{2}}) as (x\sqrt{x}).

Correct answer: A. \(\frac{2657}{784}\)

Explanation: Here \(\left(\frac{4}{7}\right)^{-2}=\frac{49}{16}\) and \(\left(\frac{7}{4}\right)^{-2}=\frac{16}{49}\). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).

Correct answer: B. \(\frac{7}{3}\)

Explanation: The left side is \(5^{2x}\cdot5^{x-2}=5^{3x-2}\), and \(3125=5^{5}\). Hence \(3x-2=5\), so \(x=\frac{7}{3}\).

Correct answer: C. (12)

Explanation: Here (\sqrt{192}=8\sqrt{3}), (2\sqrt{48}=8\sqrt{3}), and (3\sqrt{12}=6\sqrt{3}). The numerator is (6\sqrt{3}), so the value is (6).

Correct answer: A. \(\frac{17}{4}\)

Explanation: The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence \(c+r+s=2-5+7=4\).

Correct answer: A. (10\sqrt{2}+4\sqrt{5})

Explanation: Here (x^{2}=7+2\sqrt{10}), so (x^{3}=17\sqrt{2}+11\sqrt{5}) and (x^{3}-7x=10\sqrt{2}+4\sqrt{5}). In exams, first find (x^{2}) and then multiply by (x).

Correct answer: A. \(\frac{x}{4}\)

Explanation: Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.

Correct answer: B. 3

Explanation: By the laws of exponents, exponents are added when like bases are multiplied and subtracted when they are divided. Thus, the exponent on the left side is \((3p-2)+(p+5)-(2p-1)=2p+4\). Therefore, \(2p+4=10\), giving \(2p=6\) and hence \(p=3\). Option C can result from mishandling the subtraction of the denominator’s exponent. In an exam, remember to change the sign of the exponent in the denominator while simplifying.

Correct answer: C. (5^{4})

Explanation: Since (25^{-2}=5^{-4}) and (125=5^{3}), the total exponent is (9-4+3-4=4). In exams, convert all terms to the same base.

Correct answer: C. (\frac{8\sqrt{34}}{9})

Explanation: Here (u^{2}-v^{2}=(u-v)(u+v)=2\sqrt{8}\cdot2\sqrt{17}=8\sqrt{34}), and (uv=9). Hence the value is (\frac{8\sqrt{34}}{9}).

Correct answer: A. (2\sqrt{24})

Explanation: The product of the denominators is (25-24=1), and the numerator becomes (2\sqrt{24}). In exams, first find the product of conjugate denominators.

Correct answer: B. (3)

Explanation: Here (6^{x+1}-6^{x}=5\cdot6^{x}=900), so (6^{x}=180), which is not a listed integral power. Therefore none of the listed options is correct.

Correct answer: A. \(p^{8}q^{-12}\)

Explanation: Using the quotient rule for exponents, \(\frac{p^{-5}q^4}{p^{-1}q^{-2}}=p^{-5-(-1)}q^{4-(-2)}=p^{-4}q^6\). Applying \((a^m)^n=a^{mn}\), we get \((p^{-4}q^6)^{-2}=p^8q^{-12}\), so option A is correct. Option B results from mishandling the signs while multiplying the exponents by the outer \(-2\). Exam tip: subtract exponents when dividing like bases, then multiply by the outside exponent.