Exponents and radicals

Recognize the structure. Multiplication adds exponents; division subtracts the denominator’s exponent.

Work it through. $\frac{{\text{x}}^{\text{3}}{\text{x}}^{\text{5}}}{{\text{x}}^{\text{2}}}$ = x³⁺⁵⁻² = x⁶.

The restriction x ≠ 0 comes from the original denominator and remains part of the equivalence.

Answer: x⁶, for x ≠ 0.

Check. At x = 2, the original is (8)(32)/4 = 64, and 2⁶ = 64.

Avoid the trap. Multiplying the exponents 3 and 5 would be appropriate for (x³)⁵, not for x³x⁵.

Recognize the structure. The outer square affects the coefficient and both variable powers.

Work it through. First expand the power, then divide:

$\frac{{\text{9a}}^{\text{4}}{\text{b}}^{\text{−2}}}{{\text{9a}}^{\text{−1}}\text{b}}$ = a⁴⁻⁽⁻¹⁾b⁻²⁻¹ = a⁵b⁻³ = $\frac{{\text{a}}^{\text{5}}}{{\text{b}}^{\text{3}}}$.

Answer: $\frac{{\text{a}}^{\text{5}}}{{\text{b}}^{\text{3}}}$, with a ≠ 0 and b ≠ 0.

Check. The 9 cancels. Dividing by a⁻¹ = 1/a adds a factor of a, explaining the fifth power.

Avoid the trap. Subtracting a negative exponent increases the exponent: 4 − (−1) = 5, not 3.

Recognize the structure. The denominator 4 means fourth root; the negative sign means reciprocal.

Work it through. The fourth root of 16 is 2. Therefore

16^−3/4 = $\frac{\text{1}}{{\text{16}}^{\text{3/4}}}$ = $\frac{\text{1}}{\text{(}$ \sqrt[\text{4}]{\text{16}}${\text{)}}^{\text{3}}}$ = $\frac{\text{1}}{{\text{2}}^{\text{3}}}$ = $\frac{\text{1}}{\text{8}}$.

Answer: C, 1/8.

Check. Since the positive base is greater than 1 and the exponent is negative, the answer must be between 0 and 1.

Avoid the trap. A negative exponent does not create a negative value. It changes the location of the power from numerator to denominator.

Recognize the structure. 81 = 3⁴, so taking the fourth root first makes the arithmetic small.

Work it through. 81^3/4 = (3⁴)^3/4 = 3³ = 27.

Equivalently, 81^3/4 = ($\sqrt[\text{4}]{\text{81}}$)³ = 3³.

Answer: 27.

Check. Raising the answer to the fourth power gives 27⁴ = 3¹², the same as 81³.

Avoid the trap. The numerator 3 and denominator 4 do not mean “multiply by 3 and divide by 4.” They specify a power and a root.

Recognize the structure. Separate the radicand into a perfect cube times what remains.

Work it through. Write 54x⁷ = 27x⁶(2x). Since 27 = 3³ and x⁶ = (x²)³,

$\sqrt[\text{3}]{{\text{54x}}^{\text{7}}}$ = 3x²$\sqrt[\text{3}]{\text{2x}}$.

An odd root can be evaluated for negative as well as positive radicands.

Answer: 3x²$\sqrt[\text{3}]{\text{2x}}$.

Check. Cubing the simplified form gives 27x⁶(2x) = 54x⁷.

Avoid the trap. Do not impose an unnecessary x ≥ 0 restriction on a cube root. Also, the extracted power is x², because 7 = 3 ⋅ 2 + 1.

Recognize the structure. All three radicands contain square factors that leave the same remaining radicand.

Work it through. $\sqrt{\text{72}}$ = 6$\sqrt{\text{2}}$, $\sqrt{\text{8}}$ = 2$\sqrt{\text{2}}$, $\sqrt{\text{18}}$ = 3$\sqrt{\text{2}}$.

So the original expression is

6$\sqrt{\text{2}}$ − 2(2$\sqrt{\text{2}}$) + 3$\sqrt{\text{2}}$ = (6 − 4 + 3)$\sqrt{\text{2}}$ = 5$\sqrt{\text{2}}$.

Answer: 5$\sqrt{\text{2}}$.

Check. Approximate values give 8.485 − 5.657 + 4.243 ≈ 7.071, matching 5$\sqrt{\text{2}}$.

Avoid the trap. The coefficient −2 multiplies the entire simplified value of $\sqrt{\text{8}}$, so its contribution is −4$\sqrt{\text{2}}$.

Recognize the structure. An even root must be nonnegative, even when x is negative.

Work it through. Factor the radicand as 25 ⋅ 2 ⋅ (x³)². Then

$\sqrt{{\text{50x}}^{\text{6}}}$ = 5$\sqrt{\text{2}}$ |x³| = 5|x|³$\sqrt{\text{2}}$.

The absolute value is essential because the question allows all real x.

Answer: B, 5|x|³$\sqrt{\text{2}}$.

Check. At x = −1, the original equals $\sqrt{\text{50}}$ > 0. Choice A would be negative, so it cannot work for every real input.

Avoid the trap. The identity $\sqrt{{\text{x}}^{\text{6}}}$ = x³ is valid when x ≥ 0, but not for unrestricted real x.

Recognize the structure. The conjugate $\sqrt{\text{5}}$ + 1 turns the denominator into a difference of squares.

Work it through. Multiply the numerator and denominator by $\sqrt{\text{5}}$ + 1:

$\frac{\text{6}}{$ \sqrt{\text{5}}$\text{− 1}}$ ⋅ $\frac{$ \sqrt{\text{5}}$\text{+ 1}}{$ \sqrt{\text{5}}$\text{+ 1}}$ = $\frac{\text{6(}$ \sqrt{\text{5}}$\text{+ 1)}}{\text{5 − 1}}$ = $\frac{\text{3(}$ \sqrt{\text{5}}$\text{+ 1)}}{\text{2}}$.

Answer: $\frac{\text{3(}$ \sqrt{\text{5}}$\text{+ 1)}}{\text{2}}$.

Check. Both forms are approximately 4.854. The multiplier is a nonzero expression divided by itself, so it equals 1.

Avoid the trap. Multiplying only the denominator changes the fraction. Squaring $\sqrt{\text{5}}$ − 1 would not remove the radical cross term.

Recognize the structure. Use a common denominator for the exponents before combining them.

Work it through.

$\frac{{\text{a}}^{\text{2/3}}{\text{a}}^{\text{5/6}}}{{\text{a}}^{\text{1/2}}}$ = a^{2/3+5/6−1/2} = a^{4/6+5/6−3/6} = a^6/6 = a.

Answer: a.

Check. For a = 64, the factors are 16, 32, and 8, and (16)(32)/8 = 64.

Avoid the trap. Adding fractions by adding numerators and denominators would produce a meaningless exponent. Exponent arithmetic is ordinary fraction arithmetic.

Recognize the structure. Both 27 and 9 are powers of 3.

Work it through. Rewrite and simplify the exponents:

(3³)^2x−1 = (3²)^x+2 3^6x−3 = 3^2x+4.

Because the common base is positive and not 1, the exponents are equal: 6x − 3 = 2x + 4. Thus 4x = 7.

Answer: x = 7/4.

Check. Both powers of 3 have exponent 15/2 when x = 7/4.

Avoid the trap. Setting 2x − 1 = x + 2 before making the bases equal ignores the different bases and gives the wrong equation.

Recognize the structure. Apply the outer sixth power first, then combine each variable separately.

Work it through. $\frac{{\text{p}}^{\text{3}}{\text{q}}^{\text{−2}}}{{\text{p}}^{\text{1}}{\text{q}}^{\text{−1}}}$ = p³⁻¹q^{−2−(−1)} = p²q⁻¹ = $\frac{{\text{p}}^{\text{2}}}{\text{q}}$.

The positivity assumptions make all fractional powers and denominator operations valid.

Answer: p²/q.

Check. The net exponent of q is −2 − (−1) = −1. The numerator has q⁻², and dividing by q⁻¹ raises the exponent by 1.

Avoid the trap. The outer exponent multiplies both inner exponents. A negative exponent in the denominator must also be subtracted with its sign intact.

Recognize the structure. 64 = 8², so the target can be expressed using the known quantity 8^x.

Work it through. 64^x+1 = 64 ⋅ 64^x = 64 ⋅ (8²)^x = 64(8^x)² = 64 ⋅ 25 = 1600.

The value of x itself is unnecessary.

Answer: 1600.

Check. 64^x = (8^x)² = 25; increasing its exponent by 1 multiplies this by 64.

Avoid the trap. The extra +1 in the exponent cannot be ignored. Also, 64^x+1 is a product 64^x ⋅ 64, not a sum 64^x + 64.

Recognize the structure. The positivity condition permits division by a^1/3.

Work it through. Rewrite the roots as powers:

a^1/2 = 3a^1/3 a^1/2−1/3 = 3 a^1/6 = 3.

Raise both sides to the sixth power: a = 3⁶ = 729.

Answer: 729.

Check. $\sqrt{\text{729}}$ = 27 and 3$\sqrt[\text{3}]{\text{729}}$ = 3 ⋅ 9 = 27.

Avoid the trap. Without the word “positive,” a = 0 would also satisfy the original equation. Dividing by a variable power requires checking that it is nonzero.

Recognize the structure. x^2/3 means ($\sqrt[\text{3}]{\text{x}}$)². The square can hide a negative cube root.

Work it through. Let u = $\sqrt[\text{3}]{\text{x}}$. Then u² = 9, so u = 3 or u = −3. Cubing gives

x = 27 or x = −27.

Answer: x = −27 and x = 27.

Check. ($\sqrt[\text{3}]{\text{27}}$)² = 3² = 9 and ($\sqrt[\text{3}]{\text{−27}}$)² = (−3)² = 9.

Avoid the trap. Blindly raising both sides to the 3/2 power and writing x = 27 loses the negative solution. The original expression is not one-to-one over all real x.

Recognize the structure. A ratio eliminates the common factor a^1/2 and leaves a power of b.

Work it through. Divide the first equation by the second. Their positive quantities are nonzero, so division is valid: b^{1/3−(−1/3)} = $\frac{\text{6}}{\text{2}}$ b^2/3 = 3.

Cube both sides to reach the exact target: b² = 3³ = 27.

Answer: 27.

Check. Multiplying the original equations gives a = 12. With b = 3$\sqrt{\text{3}}$, a^1/2 = 2$\sqrt{\text{3}}$ and b^1/3 = $\sqrt{\text{3}}$, giving the required products 6 and 2.

Avoid the trap. Finding b first is valid but unnecessary. The question asks for b², which is obtained directly without carrying a radical into a second step.