Exponents

By the zero-exponent rule, \(x^0=1\) for a non-zero value of \(x\), and \(x^1=x\). Therefore, \(x^0+x^1=1+x\). Option D, \(x+2\), is incorrect because it adds 1 twice. Exam tip: remember that \(x^0=1\) only when \(x\ne0\).

For powers with the same exponent, use the rule \(a^n \times b^n=(ab)^n\). Thus, \(2^3 \times 5^3=(2\times5)^3=10^3\). Option B incorrectly adds the bases, while options C and D use incorrect operations. Exam tip: when the exponents are equal, multiply the bases and retain the common exponent.

For \(b\neq 0\), the exponent rule \(\frac{a^n}{b^n}=\left(\frac{a}{b}\right)^n\) applies when the numerator and denominator have the same exponent. Therefore, \(\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2\). Option D is incorrect because it does not apply the exponent 2 to the denominator. Exam tip: equal powers in the numerator and denominator can be written as the same power of the entire fraction.

By the law of exponents, \((xyz)^n=x^n y^n z^n\). Therefore, \((xyz)^2=x^2y^2z^2\), so option A is correct. In option B, the exponent has been applied only to z, whereas it must apply to every factor. Exam tip: when a power is applied to a product, apply the same power to each factor.

Use the negative-exponent rule \(a^{-n}=\frac{1}{a^n}\). Thus, \(4^{-1}=\frac{1}{4}\) and \(2^{-1}=\frac{1}{2}\). Taking a common denominator gives \(\frac{1}{4}+\frac{2}{4}=\frac{3}{4}\), so option A is correct. \(\frac{1}{8}\) would result from multiplying the two fractions, not adding them. Exam tip: convert negative exponents to reciprocals before performing the operation.

Using the negative-exponent rule, \(b^{-2}=\frac{1}{b^2}\), for \(b\neq0\). Hence, \(a^2b^{-2}=a^2\times\frac{1}{b^2}=\frac{a^2}{b^2}\). Option B reverses the numerator and denominator, while option C incorrectly changes the negative exponent to a positive one without taking the reciprocal. Exam tip: move a factor with a negative exponent across the fraction bar and change the exponent to positive.

When powers with the same base are divided, their exponents are subtracted: \(\frac{2^{-3}}{2^{-1}}=2^{-3-(-1)}=2^{-2}\). Therefore, option A is correct. Remember that a negative exponent represents a reciprocal, so \(2^{-2}=\frac{1}{4}\); \(2^2\) is a different value with a positive exponent. Exam tip: subtract the second exponent with its sign included.

The zero-exponent rule states that the zeroth power of any non-zero number or expression is 1. Therefore, for x ≠ 0, \((x^3)^0 = 1\). Option C is incorrect because a zeroth power is not generally 0. In an exam, remember that the base must be non-zero.

When the denominator is non-zero, that is, \(a \ne 0\), a non-zero expression divided by itself equals \(1\). Therefore, \(\frac{a^m}{a^m}=1\). Option B is just \(a^m\), not the value of the quotient. Exam tip: Any non-zero expression divided by itself is 1.

Since \(32 = 2^5\), the equation becomes \(2^x = 2^5\). For powers with the same base, the exponents are equal, so \(x=5\). Option 4 is incorrect because \(2^4=16\). In exams, first express the given number as a power of the same base.

Since 81 can be expressed as \(3^4\), the equation becomes \(3^x = 3^4\). For equal positive bases other than 1, their exponents must be equal; therefore, \(x=4\). Options 3 and 5 would give \(27\) and \(243\), respectively, so they are incorrect. Exam tip: rewrite the given number as a power of the same base before comparing exponents.

\(\sqrt{a^6}=\sqrt{(a^3)^2}=|a^3|\), because the square root of a number squared is its absolute value. The expression \(a^3\) is valid directly only when \(a\geq 0\) is assumed. Exam tip: retain the absolute-value sign when simplifying the square root of an even power unless the variable's sign is specified.

Use the power-of-a-power rule ((a^m)^n=a^{mn}) . Thus, ((x^{1/2})^4=x^{(1/2)\times4}=x^2) , so option A is correct. Option B leaves the original exponent unchanged, while options C and D result from multiplying the exponents incorrectly. Exam tip: when a power is raised to another power, multiply the two exponents.

If a and b have values such that a^2b^3 ≠ 0, the zero-exponent rule applies: for every non-zero number or expression, x^0 = 1. Therefore, ((a^2 b^3)^0) = 1. Option 0 is not the value of a non-zero expression raised to the zero power; it relates to a different case in which the base is zero. Exam tip: check that the base is non-zero, then use the zero-exponent rule directly.

The zeroth power of every non-zero number is 1. Thus, \(5^0=1\), \(2^0=1\), and \(3^0=1\). Therefore, \(\frac{5^0+2^0}{3^0}=\frac{1+1}{1}=2\), so option A is correct. Exam tip: remember that \(a^0=1\) for \(a\neq0\); do not confuse it with the undefined expression \(0^0\).

Using the power-of-a-power rule, \((x^2)^3=x^{2\times3}=x^6\). Then, for the product of powers with the same base, \(x^m\times x^n=x^{m+n}\), so \(x^6\times x^{-1}=x^{6+(-1)}=x^5\). Therefore, the correct answer is x^5. Exam tip: when adding exponents, include the negative sign; treating
\(x^{-1}\) as \(x^1\) would incorrectly give x^7.

When dividing powers with the same base, subtract the exponents: \(a^{3-(-1)}=a^4\) and \(b^{-2-1}=b^{-3}\). Therefore, the value is \(a^4b^{-3}\), so option A is correct. Option B results from an incorrect subtraction of the exponents. Remember: \(x^m/x^n=x^{m-n}\), with the denominator required to be non-zero.

Since 4 = 2^2, we have \(4^{-1} = (2^2)^{-1} = 2^{-2}\). Therefore, \(2^3 \cdot 4^{-1} = 2^3 \cdot 2^{-2} = 2^{3-2} = 2\). Hence, option A is correct. Exam tip: When multiplying powers with the same base, subtract the exponents; you can also verify a negative exponent by taking the reciprocal.

First, \(y^{-3}=\frac{1}{y^3}\), so \(x^3y^{-3}=\frac{x^3}{y^3}\). The exponent \(-1\) on the entire expression takes the reciprocal: \(\left(\frac{x^3}{y^3}\right)^{-1}=\frac{y^3}{x^3}\). Therefore, option A is correct. Here, \(x\) and \(y\) are assumed to be nonzero. Exam tip: remember \(a^{-n}=\frac{1}{a^n}\) and \(\left(\frac{a}{b}\right)^{-1}=\frac{b}{a}\).

In the given expression, \(a^{-2}\) is in the denominator. Using the rule \(a^{-n}=\frac{1}{a^n}\), we get \(\frac{1}{a^{-2}b^3}=\frac{a^2}{b^3}\). Therefore, option A is correct. Exam tip: when a factor with a negative exponent moves across a fraction bar, the sign of its exponent changes, while \(b^3\) remains in the denominator.