ACT Math Exponents and Roots: 24 Practice Problems with Step-by-Step Explanations

Multiply the coefficients: 2 × 9 = 18. Add the exponents of the matching base: x⁴ × x⁹ = x⁴⁺⁹ = x¹³.

Combined: 18x¹³.

Time = distance / rate = (10 × 10¹³) / (2.5 × 10⁶). Divide the coefficients: 10 / 2.5 = 4. Subtract the exponents: 10¹³⁻⁶ = 10⁷.

Combined: 4.0 × 10⁷ seconds.

A fractional exponent translates to a radical: x^a/b = ᵇ√(x^a), where the denominator becomes the root's index and the numerator stays as the power inside.

For x^3/4, the numerator is 3 and the denominator is 4, so this equals the 4th root of x³: ⁴√(x³).

The inequality 25 < a² < 49 means a² must fall strictly between 25 and 49. Test each choice by squaring it: 4.62² ≈ 21.3 (too small), 6.49² ≈ 42.1 (fits), 7² = 49 (excluded, not strictly less than 49), and 26.4² is far too large.

Only 6.49 has a square strictly between 25 and 49.

Divide the coefficients: −24/6 = −4. Subtract exponents for each base: x³⁻² = x¹, y⁶⁻¹ = y⁵.

Combined: −4xy⁵.

When multiplying powers with the same base, add the exponents: a³⁺⁵ = a⁸, b⁴⁺³ = b⁷, c²⁺⁶ = c⁸.

Combined: a⁸b⁷c⁸.

Since 7² = 49 and 8² = 64, and 49 < 52 < 64, √52 is between 7 and 8 (approximately 7.21).

The smallest integer greater than 7.21 is 8.

Rewrite 16 as 4² so every term shares base 4: (4²)^x/4³ = 4^2x/4³ = 4^2x−3.

Set the exponents equal: 2x − 3 = 7, so 2x = 10 and x = 5.

Add √2 to both sides: 5x = 10 + √2. Divide both sides by 5: x = (10 + √2)/5 = 10/5 + √2/5 = 2 + √2/5.

Divide the coefficients: 3/6 = 1/2. Subtract exponents for each base: a⁷⁻² = a⁵, c⁵⁻¹ = c⁴. Since b's exponent is larger in the denominator (5 vs. 2), it stays in the denominator as b⁵⁻² = b³.

Combined: a⁵c⁴/(2b³).

Evaluate each fractional exponent first: 25^1/2 = √25 = 5, and 8^1/3 = ³√8 = 2.

Subtract: 5 − 2 = 3. Then cube the result: 3³ = 27.

Notice that x^10y = (x^5y)², since multiplying the exponent 5y by 2 gives 10y.

Since x^5y = 10, squaring both sides gives x^10y = 10² = 100.

Expand both parts: (4x⁴)² = 16x⁸, and (2x²)⁴ = 16x⁸. The expression simplifies to 16x⁸/16x⁸ = 1 for any x that doesn't make the denominator zero.

This holds for every nonzero real number — including every nonzero integer — so there are infinitely many integer solutions (all integers except 0, where the expression is undefined).

First expand the numerator: (4a⁻³b²)³ = 4³a⁻⁹b⁶ = 64a⁻⁹b⁶.

Now divide by 16ab⁻⁴: coefficients give 64/16 = 4; for a, a⁻⁹/a¹ = a⁻¹⁰; for b, b⁶/b⁻⁴ = b¹⁰. Combined: 4a⁻¹⁰b¹⁰ = 4b¹⁰/a¹⁰.

To rationalize a binomial denominator, multiply top and bottom by its conjugate. The conjugate of (3 − √5) is (3 + √5): [5(3+√5)] / [(3−√5)(3+√5)].

The denominator becomes a difference of squares: 3² − (√5)² = 9 − 5 = 4. The numerator becomes 15 + 5√5.

Result: (15 + 5√5)/4.

Rationalize each term separately. 5/√2 = (5√2)/2 (multiplying top and bottom by √2). 3/√6 = (3√6)/6 = √6/2 (multiplying top and bottom by √6, then simplifying 3/6 to 1/2).

Both terms now share denominator 2: (5√2)/2 + √6/2 = (5√2 + √6)/2.

Split off every perfect-cube factor. 24 = 8 × 3, and 8 is a perfect cube (2³). x⁴ = x³ × x, and x³ is a perfect cube. z³ is already a perfect cube.

Pull out the perfect cubes (2, x, and z), leaving the rest under the radical: 2xz ³√(3xy²).

Expand each factor separately. (2³√6)⁶ = 2⁶ × (6^1/3)⁶ = 2⁶ × 6². (4√6)⁶ = 4⁶ × 6³ = (2²)⁶ × 6³ = 2¹² × 6³.

Multiply the two results: 2⁶ × 6² × 2¹² × 6³ = 2⁶⁺¹² × 6²⁺³ = 2¹⁸ × 6⁵.

Convert the original expression to a single exponent: ⁴√(³√(x²)) = (x²)^{1/3 × 1/4} = x^2/12 = x^1/6.

Check each option the same way: A) (x²)^1/12 = x^1/6 ✓. B) ((x²)^1/6)^1/2 = x^1/6 ✓. D) (x^1/3)^1/2 = x^1/6 ✓. But C) ⁹√x = x^1/9, which does not match the exponent 1/6 — this is the one that's NOT equivalent.

Rewrite √(4x/y) = 2√(x/y), then put both terms over a common denominator of √(xy). Multiplying √(x/y) by √(x)/√(x) gives x/√(xy), so 2√(x/y) = 2x/√(xy). Similarly, √(y/x) = y/√(xy).

Subtract over the common denominator: 2x/√(xy) − y/√(xy) = (2x−y)/√(xy).

Rewrite 3^a−2 using the quotient rule: 3^a−2 = 3^a/3² = 3^a/9. So the equation becomes 3^a/9 = 5b.

Multiply both sides by 9: 3^a = 45b.

Rewrite both sides using base 2. Since 4 = 2², ³√4 = 4^1/3 = 2^2/3, so the left side is (2^2/3)^p = 2^2p/3. Since 16 = 2⁴, the right side is (2⁴)^3q = 2^12q.

Set the exponents equal: 2p/3 = 12q. Solve for q/p by dividing both sides by 12p: q/p = (2/3)/12 = 2/36 = 1/18.

First simplify the inner fraction: x¹²/x⁵ = x^12\u22125 = x⁷.

Dividing by 1/x³ is the same as multiplying by its reciprocal, x³: x⁷ \u00d7 x³ = x⁷⁺³ = x¹⁰.

A common error is subtracting the final exponent instead of adding it (mistakenly treating "divide by 1/x³" the same as "divide by x³"), which lands on the trap answer x⁴.

For an even-indexed root of an even power, \u2074\u221a(x⁴) always equals |x| (the nonnegative value), regardless of the sign of x \u2014 this is different from an odd-indexed root, which would just return x directly with no sign correction needed.

So the equation becomes |x| = \u2212x. This is true exactly when x is negative or zero, because that's precisely when \u2212x is nonnegative and therefore able to equal the nonnegative quantity |x|. For positive x, |x| = x, which would require x = \u2212x, true only at x = 0. Combining both cases, the equation holds for all x \u2264 0.