Math · College algebra · Worked example

Simplify an expression with negative exponents

Simplify (3x⁻²)² · x⁵ / (9x⁻³), and say where the result agrees with the original.

\frac{{(3 x^{- 2})}^{2} \cdot x^{5}}{9 x^{- 3}}

Raise the product to the power

The square applies to both factors: 3² = 9, and (x⁻²)² = x⁻⁴ because the exponents multiply.

{(3 x^{- 2})}^{2} = 3^{2} x^{- 4} = 9 x^{- 4}

Multiply powers of x

Add the exponents: x⁻⁴ · x⁵ = x⁻⁴⁺⁵ = x¹.

\frac{9 x^{- 4} \cdot x^{5}}{9 x^{- 3}} = \frac{9 x}{9 x^{- 3}}

Divide

The 9s cancel, and dividing subtracts exponents: x¹ ÷ x⁻³ = x^(1 − (−3)) = x⁴.

\frac{9 x}{9 x^{- 3}} = x^{1 - (- 3)} = x^{4}

Note where it holds

The original divides by x at 0 (through x⁻² and x⁻³), so it is undefined there, while x⁴ is not. The two agree for every x ≠ 0.

Check with a number

At x = 2 the original is (3/4)² · 32 ÷ (9/8) = 18 ÷ 9/8 = 16, and 2⁴ = 16.

Result

The expression simplifies to x⁴, for x ≠ 0.

x^{4} (x \neq 0)

Your turn

Evaluate 27^(2/3) and 16^(−3/4).

Show the answer and explanation

9 and 1/8.

27^(2/3) = (∛27)² = 3² = 9. 16^(−3/4) = 1/16^(3/4) = 1/(⁴√16)³ = 1/2³ = 1/8.

\begin{array}{l} 27^{\frac{2}{3}} = 9 \\ 16^{- \frac{3}{4}} = \frac{1}{8} \end{array}

Keep exploring

Open the expression in Steps, assumptions & inequalities and clear the interval: without x > 0 the step is not established, because the domains differ at 0.

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