Exponent and Radical Rules

One Set of Rules for Powers and Roots

Exponents and radicals are the same idea written two ways. Once you accept that a radical is just a fractional exponent — $\sqrt{x} = x^{1/2}$ — every root question becomes an exponent question, and one small set of laws handles all of them. The SAT rewards students who move fluidly between $\sqrt[3]{x^2}$ and $x^{2/3}$ without hesitation, because the "hard" problems are usually just two or three easy rules stacked together.

The Core Exponent Laws

For any nonzero base and real exponents:

Rule	Statement	Example
Product	$x^m \cdot x^n = x^{m+n}$	$x^3 \cdot x^4 = x^7$
Quotient	$\dfrac{x^m}{x^n} = x^{m-n}$	$\dfrac{x^7}{x^2} = x^5$
Power of a power	$(x^m)^n = x^{mn}$	$(x^3)^2 = x^6$
Power of a product	$(xy)^n = x^n y^n$	$(2x)^3 = 8x^3$
Zero exponent	$x^0 = 1$	$9^0 = 1$
Negative exponent	$x^{-n} = \dfrac{1}{x^n}$	$2^{-3} = \tfrac{1}{8}$

The two rules students confuse most are the first and third. When you multiply like bases you add exponents; when you raise a power to a power you multiply exponents. Say it out loud each time — "multiply, so add" — until it is automatic.

Negative and Fractional Exponents

A negative exponent means "reciprocal," not "negative number." So $5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}$, a positive value. Moving a factor across a fraction bar flips the sign of its exponent: $\dfrac{1}{x^{-3}} = x^3$.

A fractional exponent encodes a root. The denominator is the root and the numerator is the power:

$x^{m/n} = \sqrt[n]{x^m} = \left(\sqrt[n]{x}\right)^m.$

For example, $8^{2/3} = \left(\sqrt[3]{8}\right)^2 = 2^2 = 4$. Taking the root first keeps the numbers small — cube-root the $8$ to get $2$ before squaring, rather than squaring $8$ to $64$ and then hunting for its cube root.

Simplifying Radicals

To simplify a square root, factor out the largest perfect square. For $\sqrt{72}$, write $72 = 36 \cdot 2$, and since $36$ is a perfect square,

$\sqrt{72} = \sqrt{36 \cdot 2} = \sqrt{36} \cdot \sqrt{2} = 6\sqrt{2}.$

The same product rule handles variables: $\sqrt{x^6} = x^3$, because $x^3 \cdot x^3 = x^6$. Radicals with the same simplified root can then be combined like like-terms: $\sqrt{72} + \sqrt{50} = 6\sqrt{2} + 5\sqrt{2} = 11\sqrt{2}$.

A Fully Worked Example

Simplify $\dfrac{(2x^3)^2 \cdot x^{-1}}{4x^2}$.

Step 1 — expand the power of a product. $(2x^3)^2 = 2^2 \cdot (x^3)^2 = 4x^6$. The exponent $2$ hits both the $2$ and the $x^3$, and $(x^3)^2 = x^6$ by the power-of-a-power rule.

Step 2 — multiply the numerator. $4x^6 \cdot x^{-1} = 4x^{6 + (-1)} = 4x^5$, adding exponents on the like bases.

Step 3 — divide.

$\frac{4x^5}{4x^2} = \frac{4}{4} \cdot x^{5-2} = 1 \cdot x^3 = x^3.$

The answer is $x^3$. Each step used exactly one law — expand, then add, then subtract — which is the discipline the SAT is really testing.

Connecting Radicals and Rational Exponents

When a question buries a variable inside nested roots, convert everything to exponents. For instance,

$\sqrt{x \cdot \sqrt{x}} = \sqrt{x \cdot x^{1/2}} = \sqrt{x^{3/2}} = x^{3/4}.$

Rewriting in exponent form turns an intimidating expression into simple arithmetic on fractions. The same trick powers the hardest items: an expression like $(16x^8 y^{12})^{3/4}$ becomes $16^{3/4} \cdot x^{6} \cdot y^{9} = 8x^6 y^9$ once you distribute the $\tfrac34$ power across each factor and evaluate $16^{3/4} = (2^4)^{3/4} = 2^3 = 8$.