Complicated Exponent Laws (Index Laws)

Complicated Exponent Laws (Index Laws)

So far, we have considered situations where one particular exponent’s law was used for simplifying expressions with exponents (indices). However, in most practical situations, more than one law is needed to simplify the expression.

The following example simplifies expressions with exponents (indices) using several exponent laws.

YouTube player

Example 1

Write $64^{\frac{2}{3}}$ in simplest form.

\( \begin{align} \displaystyle
64^{\frac{2}{3}} &= (4^3)^{\frac{2}{3}} \\
&= 4^{3 \times \frac{2}{3}} \\
&= 4^2 \\
&= 16
\end{align} \)

When more than one exponent law (index law) is used to simplify an expression, the following steps can be taken.

$\textit{Step 1}$: If an expression contains brackets, expand them first.
$\textit{Step 2}$: If an expression is a fraction, simplify each numerator and denominator, then divide (simplify across then down).
$\textit{Step 3}$: Express the final answer with positive exponents (indices).

The following examples illustrate the use of exponent laws (index law) for the multiplication and division of fractions.

YouTube player

Example 2

Simplify $\dfrac{(2x^2y^3)^3 \times 3(xy^4)^2}{6x^4 \times 2xy^4}$.

\( \begin{align} \displaystyle
\dfrac{(2x^2y^3)^3 \times 3(xy^4)^2}{6x^4 \times 2xy^4} &= \dfrac{2^3x^6y^9 \times 3x^2y^8}{(6 \times 2) \times x^{4+1}y^4} \\
&= \dfrac{(8 \times 3) \times x^{6+2} \times y^{9+8}}{12x^5y^4} \\
&= \dfrac{24x^8y^{17}}{12x^5y^4} \\
&= \dfrac{24}{12} \times x^{8-5} \times y^{17-4} \\
&= 2x^3y^{13}
\end{align} \)

YouTube player

Example 3

Simplify $a^{-2}b^4 \times (a^3b^{-4})^{-1}$, leaving your answer with positive exponents.

\( \begin{align} \displaystyle
a^{-2}b^4 \times (a^3b^{-4})^{-1} &= a^{-2}b^4 \times a^{-3}b^4 \\
&= a^{-2-3}b^{4+4} \\
&= a^{-5}b^8 \\
&= \dfrac{b^8}{a^5}
\end{align} \)

YouTube player

Example 4

Simplify $\Bigg(\dfrac{a^{-\frac{1}{2}}b^{-1}}{3^{-1}b^2}\Bigg)^{-1} \div \Bigg(\dfrac{3a^{-\frac{3}{2}}b^2}{a^{\frac{3}{4}}b^{\frac{1}{2}}}\Bigg)^2$, leaving your answer with positive exponents.

\( \begin{align} \displaystyle
\Bigg(\dfrac{a^{-\frac{1}{2}}b^{-1}}{3^{-1}b^2}\Bigg)^{-1} \div \Bigg(\dfrac{3a^{-\frac{3}{2}}b^2}{a^{\frac{3}{4}}b^{\frac{1}{2}}}\Bigg)^2 &= \dfrac{a^{\frac{1}{2}}b}{3b^{-2}} \div \dfrac{3^2a^{-3}b^4}{a^{\frac{3}{2}}b} \\
&= \dfrac{a^{\frac{1}{2}}b}{3b^{-2}} \times \dfrac{a^{\frac{3}{2}}b}{3^2a^{-3}b^4} \\
&= \dfrac{a^2b^2}{27a^{-3}b^2} \\
&= \dfrac{a^5}{27}
\end{align} \)

YouTube player

Example 5

Simplify $\dfrac{3^n \times 6^{n+1} \times 12^{n-1}}{3^{2n} \times 8^n}$.

\( \begin{align} \displaystyle
\dfrac{3^n \times 6^{n+1} \times 12^{n-1}}{3^{2n} \times 8^n} &= \dfrac{3^n \times (3 \times 2)^{n+1} \times (2^2 \times 3)^{n-1}}{3^{2n} \times 2^{3n}} \\
&= \dfrac{3^n \times 3^{n+1} \times 2^{n+1} \times 2^{2n-2} \times 3^{n-1}}{3^{2n} \times 2^{3n}} \\
&= \dfrac{3^{3n} \times 2^{3n-1}}{3^{2n} \times 2^{3n}} \\
&= 3^n \times 2^{-1} \\
&= \dfrac{3^n}{2}
\end{align} \)


Algebra Algebraic Fractions Arc Binomial Expansion Capacity Chain Rule Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Pythagoras Theorem Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume

Related Articles


Your email address will not be published. Required fields are marked *