• Product Law: \(a^m\times a^n=a^{m+n}\)
• Quotient Law: \(\dfrac{a^m}{a^n}=a^{m-n}\) (for \(a\neq 0\))
• Power Law: \((a^m)^n=a^{mn}\)
• Zero Exponent: \(a^0=1\) (for \(a\neq 0\))
• Negative Exponent: \(a^{-n}=\dfrac{1}{a^n}\) (for \(a\neq 0\))

• \(a^{\frac{1}{n}}\) represent the \(n\)-th root of \(a:a^{\frac{1}{n}}=\sqrt[n]{a}\)
• \(a^\frac{m}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m\)

• Combine same terms using the laws of indices.
• Simplify expressions with indices by applying the appropriate laws.

• Base: The number that is multiplied.
• Exponent: The number of times the base is multiplied by itself.
• Index Form: Expression written with exponent such as \(2^3\).

Simplify the following algebraic expression.

\((5x^{-1})^3\times4xy^2 \div (xy)^{-4}\)

\(\begin{aligned} &(5x^{-1})^3\times4xy^2 \div (xy)^{-4} \\ &=\dfrac{(5x^{-1})^3\times4xy^2}{(xy)^{-4}} \\ &=5^3x^{-3}\times 4xy^2 \times (xy)^4 \\ &=125\times 4\times x^{-3+1+4}\times y^{2+4} \\ &=500x^2y^6. \end{aligned}\)

Show that \(7^{2x-1}=\dfrac{49^x}{7}\).

Solve for the equation on the left hand side.

\(\begin{aligned} 7^{2x-1}&=\dfrac{7^{2x}}{7} \\ &=\dfrac{49^x}{7}. \end{aligned}\)

Solve the following equation.

\(32^x=\dfrac{1}{8^{x-1}}\)

\(\begin{aligned} 32^x&=\dfrac{1}{8^{x-1}} \\ 2^{5x}&=2^{-3(x-1)} \\ 5x&=-3x+3 \\ 8x&=3 \\ x&=\dfrac{3}{8}. \end{aligned}\)

• Product Law: \(a^m\times a^n=a^{m+n}\)
• Quotient Law: \(\dfrac{a^m}{a^n}=a^{m-n}\) (for \(a\neq 0\))
• Power Law: \((a^m)^n=a^{mn}\)
• Zero Exponent: \(a^0=1\) (for \(a\neq 0\))
• Negative Exponent: \(a^{-n}=\dfrac{1}{a^n}\) (for \(a\neq 0\))

• \(a^{\frac{1}{n}}\) represent the \(n\)-th root of \(a:a^{\frac{1}{n}}=\sqrt[n]{a}\)
• \(a^\frac{m}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m\)

• Combine same terms using the laws of indices.
• Simplify expressions with indices by applying the appropriate laws.

• Base: The number that is multiplied.
• Exponent: The number of times the base is multiplied by itself.
• Index Form: Expression written with exponent such as \(2^3\).

Simplify the following algebraic expression.

\((5x^{-1})^3\times4xy^2 \div (xy)^{-4}\)

\(\begin{aligned} &(5x^{-1})^3\times4xy^2 \div (xy)^{-4} \\ &=\dfrac{(5x^{-1})^3\times4xy^2}{(xy)^{-4}} \\ &=5^3x^{-3}\times 4xy^2 \times (xy)^4 \\ &=125\times 4\times x^{-3+1+4}\times y^{2+4} \\ &=500x^2y^6. \end{aligned}\)

Show that \(7^{2x-1}=\dfrac{49^x}{7}\).

Solve for the equation on the left hand side.

\(\begin{aligned} 7^{2x-1}&=\dfrac{7^{2x}}{7} \\ &=\dfrac{49^x}{7}. \end{aligned}\)

Solve the following equation.

\(32^x=\dfrac{1}{8^{x-1}}\)

\(\begin{aligned} 32^x&=\dfrac{1}{8^{x-1}} \\ 2^{5x}&=2^{-3(x-1)} \\ 5x&=-3x+3 \\ 8x&=3 \\ x&=\dfrac{3}{8}. \end{aligned}\)