6.1: Exponents rules and properties (2024)

Simplify: \(a^3\cdot a^2\)

Solution

First, let’s rewrite this product in expanded form and then combine with one base \(a\).

\[\begin{array}{rl}a^3\cdot a^2&\text{Expand} \\ (a\cdot a\cdot a)\cdot (a\cdot a)&\text{Rewrite with one base }a \\ \underset{\color{blue}{5\text{ times}}}{\underbrace{a\cdot a\cdot a\cdot a\cdot a}}&\text{Multiplying }a\text{ five times} \\ a^5&\text{Simplified expression}\end{array}\nonumber\]

Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[a^n\cdot a^m=a^{n+m}\nonumber\]

Simplify: \(3^2\cdot 3^6\cdot 3\)

Solution

Let’s apply the product rule and simplify. Don’t forget that \(3\) has an exponent, it is one: \(3^1\).

We don’t always write it, but we know it’s there.

\[\begin{array}{rl}3^2\cdot 3^6\cdot 3^{\color{blue}{1}}\color{black}{}&\text{Same base} \\ 3^{2+6+1}&\text{Add the exponents} \\ 3^9&\text{Simplified expression}\end{array}\nonumber\]

We can simplify this even more as \(19,683\) \((3^9 = 19683)\).

Simplify: \((2x^3y^5z)\cdot (5xy^2z^3)\)

Solution

\[\begin{array}{rl}(2x^3y^5z)\cdot (5xy^2z^3)&\text{Rewrite without parenthesis} \\ 2x^3y^5z^{\color{blue}{1}}\color{black}{}\cdot 5x^{\color{blue}{1}}\color{black}{}y^2z^3&\text{Multiply the coefficients and add exponents with same bases} \\ 2\cdot 5\cdot x^{3+1}\cdot y^{5+2}\cdot z^{1+3}&\text{Add exponents and multiply the coefficients} \\ 10x^4y^7z^4&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \(\dfrac{a^5}{a^2}\)

Solution

\[\begin{array}{rl}\dfrac{a^5}{a^2}&\text{Expand} \\ \dfrac{a\cdot a\cdot a\cdot a\cdot a}{a\cdot a}&\text{Reduce the common factors} \\ \dfrac{\cancel{a}\cdot \cancel{a}\cdot a\cdot a\cdot a}{\cancel{a}\cdot\cancel{a}}&\text{Simplify} \\ a\cdot a\cdot a&\text{Rewrite with one base }a \\ a^3&\text{Simplified expression}\end{array}\nonumber\]

Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[\dfrac{a^n}{a^m}=a^{n-m}\nonumber\]

Simplify: \(\dfrac{7^{13}}{7^5}\)

Solution

\[\begin{array}{rl}\dfrac{7^{13}}{7^5}&\text{Same base} \\ 7^{13-5}&\text{Subtract exponents} \\ 7^8&\text{Simplified expression}\end{array}\nonumber\]

We can simplify this even more as \(5,764,801\) (\(7^8 = 5764801\)).

Simplify: \(\dfrac{5a^3b^5c^2}{2ab^3c}\)

Solution

\[\begin{array}{rl}\dfrac{5a^3b^5c^2}{2a^{\color{blue}{1}}\color{black}{}b^3c^{\color{blue}{1}}\color{black}{}}&\text{Subtract exponents with same bases and simplify coefficients, if possible} \\ \dfrac{5a^{3-1}b^{5-3}c^{2-1}}{2}&\text{Simplify} \\ \dfrac{5a^2b^2c}{2}&\text{Simplified expression}\end{array}\nonumber\]

We could also write the expression with the fraction as a coefficient: \(\dfrac{5}{2} a^2 b^2 c\). These are equivalent and both correct.

Simplify: \((a^2)^3\)

Solution

First, let’s rewrite this expression in expanded form and then combine with one base \(a\).

\[\begin{array}{rl}(a^2)^3&\text{Expand} \\ a^2\cdot a^2\cdot a^2&\text{Apply the product rule} \\ a^{2+2+2}&\text{Add exponents} \\ a^6&\text{Simplified expression}\end{array}\nonumber\]

Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[(a^n)^m=a^{m\cdot n}\nonumber\]

Simplify: \((ab)^3\)

Solution

We can expand the base, then rewrite with one base of \(a\) and \(b\).

\[\begin{array}{rl}(ab)^3&\text{Expand} \\ (ab)(ab)(ab)&\text{Let's rewrite this grouping }a\text{'s and }b\text{'s} \\ a\cdot a\cdot a\cdot b\cdot b\cdot b&\text{Rewrite with one base of }a\text{ and }b\\ a^3b^3&\text{Simplified expression}\end{array}\nonumber\]

Let \(a\) and \(b\) be a positive real numbers and \(n\) be any real number. Then \[(ab)^n=a^n\cdot b^m\nonumber\]

It is important to be careful to only use the power of a product rule with multiplication inside parenthesis. This property is not allowed for addition or subtraction, i.e., \[(a+b)^m\neq a^m+b^m\nonumber\]

Simplify: \(\left(\dfrac{a}{b}\right)^3\)

Solution

Let’s expand the fraction and rewrite with one base of \(a\) and \(b\).

\[\begin{array}{rl}\left(\dfrac{a}{b}\right)^3&\text{Expand} \\ \left(\dfrac{a}{b}\right)\left(\dfrac{a}{b}\right)\left(\dfrac{a}{b}\right)&\text{Multiply fractions} \\ \dfrac{a^3}{b^3}&\text{Simplified expression}\end{array}\nonumber\]

Let \(a\) and \(b\) be a positive real numbers and \(n\) be any real number. Then \[\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\nonumber\]

Simplify: \((x^3yz^2)^4\)

Solution

\[\begin{array}{rl}(x^3y^{\color{blue}{1}}\color{black}{}z^2)^4&\text{Apply the POP rule} \\ x^{3\cdot 4}y^{1\cdot 4}z^{2\cdot 4}&\text{Multiply exponents} \\ x^{12}y^4z^8&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \(\left(\dfrac{a^3b}{c^8d^5}\right)^2\)

Solution

\[\begin{array}{rl}\left(\dfrac{a^3b^{\color{blue}{1}}\color{black}{}}{c^8d^5}\right)^2&\text{Apply the power of a quotient rule} \\ \dfrac{a^{3\cdot 2}b^{1\cdot 2}}{c^{8\cdot 2}d^{5\cdot 2}}&\text{Multiply exponents} \\ \dfrac{a^6b^2}{c^{16}d^{10}}&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \((4x^2y^5)^3\)

Solution

\[\begin{array}{rl}(4^{\color{blue}{1}}\color{black}{}x^2y^5)^3&\text{Apply the POP rule} \\ 4^{3\cdot 1}x^{2\cdot 3}y^{5\cdot 3}&\text{Multiply exponents} \\ 4^3\cdot x^6\cdot y^{15}&\text{Evaluate }4^3 \\ 64x^6y^{15}&\text{Simplified expression}\end{array}\nonumber\]

Notice that the exponent also applied to the coefficient \(4\) and we had to evaluate \(4^3 = 64\) as part of the expression.

Let \(a\) and \(b\) be positive real numbers and \(n\) and \(m\) be any real numbers.

Rule 1. \(a^n\cdot a^m=a^{n+m}\)

Rule 2. \(\dfrac{a^n}{a^m}=a^{n-m}\)

Rule 3. \((a^n)^m=a^{nm}\)

Rule 4. \((ab)^n=a^n\cdot b^n\)

Rule 5. \(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)

Simplify: \(\dfrac{a^3}{a^3}\)

Solution

If we applied the quotient rule right away, we would get

\[\begin{aligned}\dfrac{a^3}{a^3}&=a^{3-3} \\ &=a^0\end{aligned}\]

But what does this mean? What is \(a^0\)? Well, let’s take a look at this same example with a different approach:

\[\begin{array}{rl}\dfrac{a^3}{a^3}&\text{Expand} \\ \dfrac{a\cdot a\cdot a}{a\cdot a\cdot a}&\text{Reduce common factors of }a \\ \dfrac{\cancel{a}\cdot\cancel{a}\cdot\cancel{a}}{\cancel{a}\cdot\cancel{a}\cdot\cancel{a}}&\text{Simplify} \\ \dfrac{1}{1}&\text{Simplify} \\ 1&\text{Simplified expression}\end{array}\nonumber\]

If \(\dfrac{a^3}{a^3}=a^0\) from the first part and \(\dfrac{a^3}{a^3}=1\) from the second part, then this implies \(a^0=1\).

Let \(a\) be a positive real number. Then \(a^0 = 1\), i.e., any positive real number to the power of zero is \(1\).

Simplify: \((3x^2)^0\)

Solution

Since \(3x^2\) is raised to the power of zero, then we can apply the zero power rule:

\[\begin{array}{rl}(3x^2)^0&\text{Zero power rule} \\ 1&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \(\dfrac{a^3}{a^5}\)

Solution

If we applied the quotient rule right away, we would get

\[\begin{aligned}\dfrac{a^3}{a^5}&=a^{3-5} \\ &=a^{-2}\end{aligned}\]

But what does this mean? What is \(a^{−2}\)? Well, let’s take a look at this same example with a different approach:

\[\begin{array}{rl}\dfrac{a^3}{a^5}&\text{Expand} \\ \dfrac{a\cdot a\cdot a}{a\cdot a\cdot a\cdot a\cdot a}&\text{Reduce common factors of }a \\ \dfrac{\cancel{a}\cdot\cancel{a}\cdot\cancel{a}}{\cancel{a}\cdot\cancel{a}\cdot\cancel{a}\cdot a\cdot a}&\text{Simplify} \\ \dfrac{1}{a\cdot a}&\text{Simplify} \\ \dfrac{1}{a^2}&\text{Simplified expession}\end{array}\nonumber\]

If \(\dfrac{a^3}{a^5}=a^{-2}\) from the first part and \(\dfrac{a^3}{a^5}=\dfrac{1}{a^2}\) from the second part, then this implies \(a^{-2}=\dfrac{1}{a^2}\).

It is important to note a negative exponent does not imply the expression is negative, only the reciprocal of the base. Hence, negative exponents imply reciprocals.

Also, recall the rules of simplifying:

• All exponents are positive
• Each base only occurs once
• There are no parenthesis
• There are no powers written to powers

This includes rewriting all negative exponents as positive exponents.

Let \(a\) and \(b\) be positive real numbers and \(n\) be any real number.

Rule 1. \(a^{-n}=\dfrac{1}{a^n}\)

Rule 2. \(\dfrac{1}{a^{-n}}=a^n\)

Rule 3. \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n\)

Simplify: \(\dfrac{a^3b^{-2}c}{2d^{-1}e^{-4}f^2}\)

Solution

We can rewrite the expression with positive exponents using the rules of exponents:

\[\begin{array}{rl}\dfrac{a^3b^{-2}c}{2d^{-1}e^{-4}f^2}&\text{Reciprocate the terms with negative exponents} \\ \dfrac{a^3cde^4}{2b^2f^2}&\text{Simplified expression}\end{array}\nonumber\]

As we simplified the fraction, we took special care to move each base that had a negative exponent, but the expression itself did not become negative. Also, it is important to remember that exponents only effect the base. The \(2\) in the denominator has an exponent of one (we don’t always write it, but we know it’s there), so it does not move with the \(d\).

Nicolas Chuquet, the French mathematician of the \(15^{\text{th}}\) century wrote \(12^{1\overline{m}}\) to indicate \(12x^{−1}\). This was the first known use of the negative exponent.

Let \(a\) and \(b\) be positive real numbers and \(n\) and \(m\) be any real numbers.

Rule 1. \(a^n\cdot a^m=a^{n+m}\)

Rule 2. \(\dfrac{a^n}{a^m}=a^{n-m}\)

Rule 3. \((a^n)^m=a^{nm}\)

Rule 4. \((ab)^n=a^n\cdot b^n\)

Rule 5. \(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)

Rule 6. \(a^0=1\)

Rule 7. \(a^{-n}=\dfrac{1}{a^n}\)

Rule 8. \(\dfrac{1}{a^{-n}}=a^n\)

Rule 9. \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n\)

Simplify: \(\dfrac{4x^{-5}y^{-3}\cdot 3x^3y^{-2}}{6x^{-5}y^3}\)

Solution

\[\begin{array}{rl}\dfrac{4x^{-5}y^{-3}\cdot 3x^3y^{-2}}{6x^{-5}y^3}&\text{Simplify the numerator by applying the product rule} \\ \dfrac{12x^{-2}y^{-5}}{6x^{-5}y^3}&\text{Simplify by applying the quotient rule} \\ \dfrac{12}{6}\cdot x^{-2-(-5)}y^{-5-3}&\text{Simplify} \\ 2x^3y^{-8}&\text{Rewrite with only positive exponents} \\ \dfrac{2x^3}{y^8}&\text{Simplified expression}\end{array}\nonumber\]

Simplify: \(\dfrac{(3ab^3)^{-2}\cdot ab^{-3}}{2a^{-4}b^0}\)

Solution

\[\begin{array}{rl}\dfrac{(3ab^3)^{-2}\cdot a^1b^{-3}}{2a^{-4}b^0}&\text{Apply POP and zero power rule} \\ \dfrac{3^{-2}a^{-2}b^{-6}\cdot a^1b^{-3}}{2a^{-4}\cdot 1}&\text{Apply product rule} \\ \dfrac{3^{-2}a^{-1}b^{-9}}{2a^{-4}}&\text{Apply the quotient rule} \\ \dfrac{3^{-2}a^3b^{-9}}{2}&\text{Rewrite with only positive exponents} \\ \dfrac{a^3}{2\cdot 3^2\cdot b^9}&\text{Simplify} \\ \dfrac{a^3}{18b^9}&\text{Simplified expression}\end{array}\nonumber\]

It is important to point out that when we simplified \(3^{−2}\), we moved the \(3^{−2}\) to the denominator and the exponent became positive. We did not make the number negative. Negative exponents never make the bases negative; they simply mean we have to take the reciprocal of the base.

Simplify: \(\left(\dfrac{3x^{-2}y^5z^3\cdot 6x^{-6}y^{-2}z^{-3}}{9(x^2y^{-2})^{-3}}\right)^{-3}\)

Solution

This example looks more involved that any of the other examples, but we will apply the same method. It is advised, in these types of problems, that we simplify the expression inside the parenthesis first, and then apply the POP rule. We even should start with simplifying each numerator and denoinator before simplifying the fraction with the quotient rule.

\[\begin{array}{rl} \left(\dfrac{3x^{-2}y^5z^3\cdot 6x^{-6}y^{-2}z^{-3}}{9(x^2y^{-2})^{-3}}\right)^{-3}&\text{Simplify each numeratr and denominator} \\ \left(\dfrac{18x^{-8}y^3z^0}{9x^{-6}y^6}\right)^{-3}&\text{Apply the quotient rule} \\ (2x^{-2}y^{-3}z^0)^{-3}&\text{Apply the POP rule} \\ 2^{-3}x^{-6}y^9z^0&\text{Rewrite only with positive exponents} \\ \dfrac{x^6y^9}{2^3}&\text{Simplify} \\ \dfrac{x^6y^9}{8}&\text{Simplified expression}\end{array}\nonumber\]

\(4\cdot 4^4\cdot 4^4\)

\(4\cdot 2^2\)

\(3m\cdot 3mn\)

\(2m^4n^2\cdot 4nm^2\)

\((3^3)^4\)

\((4^4)^2\)

\((2u^3v^2)^2\)

\((2a^4)^4\)

\(\dfrac{4^5}{4^3}\)

\(\dfrac{3^2}{3}\)

\(\dfrac{3nm^2}{3n}\)

\(\dfrac{4x^3y^4}{3xy^3}\)

\((x^3y^4\cdot 2x^2y^3)^2\)

\(2x(x^4y^4)^4\)

\(\dfrac{2x^7y^5}{3x^3y\cdot 4x^2y^3}\)

\(\left(\dfrac{(2x)^3}{x^3}\right)^2\)

\(\left(\dfrac{2y^{17}}{(2x^2y^4)^4}\right)^3\)

\(\left(\dfrac{2mn^4\cdot 2m^4n^4}{mn^4}\right)^3\)

\(\dfrac{2xy^5\cdot 2x^2y^3}{2xy^4\cdot y^3}\)

\(\dfrac{q^3r^2\cdot (2p^2q^2r^3)^2}{2p^3}\)

\(\left(\dfrac{zy^3\cdot z^3x^4y^4}{x^3y^3z^3}\right)^4\)

\(\dfrac{2x^2y^2z^6\cdot 2zx^2y^2}{(x^2z^3)^2}\)

\(4\cdot 4^4\cdot 4^2\)

\(3\cdot 3^3\cdot 3^2\)

\(3x\cdot 4x^2\)

\(x^2y^4\cdot xy^2\)

\((4^3)^4\)

\((3^2)^3\)

\((xy)^3\)

\((2xy)^4\)

\(\dfrac{3^7}{3^3}\)

\(\dfrac{3^4}{3}\)

\(\dfrac{x^2y^4}{4xy}\)

\(\dfrac{xy^3}{4xy}\)

\((u^2v^2\cdot 2u^4)^3\)

\(\dfrac{3vu^5\cdot 2v^3}{uv^2\cdot 2u^3v}\)

\(\dfrac{2ba^7\cdot 2b^4}{ba^2\cdot 3a^3b^4}\)

\(\dfrac{2a^2b^2a^7}{(ba^4)^2}\)

\(\dfrac{yx^2\cdot (y^4)^2}{2y^4}\)

\(\dfrac{n^3(n^4)^2}{2mn}\)

\(\dfrac{(2y^3x^2)^2}{2x^2y^4\cdot x^2}\)

\(\dfrac{2x^4y^5\cdot 2z^{10}x^2y^7}{(xy^2z^2)^4}\)

\(\left(\dfrac{2q^3p^3r^4\cdot 2p^3}{(qrp^3)^2}\right)^4\)

\(2x^4y^{-2}\cdot (2xy^3)^4\)

\((a^4b^{-3})^3\cdot 2a^3b^{-2}\)

\((2x^2y^2)^4x^{-4}\)

\((x^3y^4)^3\cdot x^{-4}y^4\)

\(\dfrac{2x^{-3}y^2}{3x^{-3}y^3\cdot 3x^0}\)

\(\dfrac{4xy^{-3}\cdot x^{-4}y^0}{4y^{-1}}\)

\(\dfrac{u^2v^{-1}}{2u^0v^4\cdot 2uv}\)

\(\dfrac{u^2}{4u^0v^3\cdot 3v^2}\)

\(\dfrac{2y}{(x^0y^2)^4}\)

\(\left(\dfrac{2a^2b^3}{a^{-1}}\right)^4\)

\(\dfrac{2nm^4}{(2m^2n^2)^4}\)

\(\dfrac{(2mn)^4}{m^0n^{-2}}\)

\(\dfrac{y^3\cdot x^{-3}y^2}{(x^4y^2)^3}\)

\(\dfrac{2u^{-2}v^3\cdot (2uv^4)^{-1}}{2u^{-4}v^0}\)

\(\left(\dfrac{2x^0\cdot y^4}{y^4}\right)^3\)

\(\dfrac{y(2x^4y^2)^2}{2x^4y^0}\)

\(\dfrac{2yzx^2}{2x^4y^4z^{-2}\cdot (zy^2)^4}\)

\(\dfrac{2kh^0\cdot 2h^{-3}k^0}{(2kj^3)^2}\)

\(\dfrac{(cb^3)^2\cdot 2a^{-3}b^2}{(a^3b^{-2}c^3)^3}\)

\(\dfrac{(yx^{-4}z^2)^{-1}}{z^3\cdot x^2y^3z^{-1}}\)

\(2a^{-2}b^{-3}\cdot (2a^0b^4)^4\)

\(2x^3y^2\cdot (2x^3)^0\)

\((m^0n^3\cdot 2m^{-3}n^{-3})^0\)

\(2m^{-1}n^{-3}\cdot (2m^{-1}n^{-3})^4\)

\(\dfrac{3y^3}{3yx^3\cdot 2x^4y^{-3}}\)

\(\dfrac{3x^3y^2}{4y^{-2}\cdot 3x^{-2}y^{-4}}\)

\(\dfrac{2xy^2\cdot 4x^3y^{-4}}{4x^{-4}y^{-4}\cdot 4x}\)

\(\dfrac{2x^{-2}y^2}{4yx^2}\)

\(\dfrac{(a^4)^4}{2b}\)

\(\left(\dfrac{2y^{-4}}{x^2}\right)^{-2}\)

\(\dfrac{2y^2}{(x^4y^0)^{-4}}\)

\(\dfrac{2x^{-3}}{(x^4y^{-3})^{-1}}\)

\(\dfrac{2x^{-2}y^0\cdot 2xy^4}{(xy^0)^{-1}}\)

\(\dfrac{2yx^2\cdot x^{-2}}{(2x^0y^4)^{-1}}\)

\(\dfrac{u^{-3}v^{-4}}{2v(2u^{-3}v^4)^0}\)

\(\dfrac{b^{-1}}{(2a^4b^0)^0\cdot 2a^{-3}b^2}\)

\(\dfrac{2b^4c^{-2}\cdot (2b^3c^2)^{-4}}{a^{-2}b^4}\)

\(\left(\dfrac{(2x^{-3}y^0z^{-1})^3\cdot x^{-3}y^2}{2x^3}\right)^{-2}\)

\(\dfrac{2q^4\cdot m^2p^2q^4}{(2m^{-4}p^2)^3}\)

\(\dfrac{2mpn^{-3}}{(m^0n^{-4}p^2)^3\cdot 2n^2p^0}\)