6.1: Exponents rules and properties
- Page ID
- 45058
If \(a\) is a positive real number and \(n\) is any real number, then in \[a^n,\nonumber\] \(a\) is called the base and \(n\) is called the exponent.
When the directions state simplify, this means
- All exponents are positive
- Each base only occurs once
- There are no parenthesis
- There are no powers written to powers
Product Rule of Exponents
Let’s take a look at an example with multiplication.
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’s think about Example 6.1.1 . This method of expanding seems to be fine when there are smaller exponents, but what if we were given something like \(a^{100}\cdot a^{934}\)? Are we going to expand \(a\) over a thousand times? No way! We need a more sophisticated way in multiplying expressions with exponents. Hence, taking a look at Example 6.1.1 , we can see the result is \(a^5\). Notice we could have obtained this answer without expanding but by simply adding the exponents:
\[a^3\cdot a^2=a^{3+2}=a^5\nonumber\]
This is called the product rule of exponents.
Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[a^n\cdot a^m=a^{n+m}\nonumber\]
In order to add exponents, the bases of the factors are required to be the same.
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
\[\begin{array}{rl}(2x^3y^5z)\cdot (5xy^2z^3)&\text{Rewrite without parenthesis} \\ 2x^3y^5z^{\color{blue}{1}}\color{black}{}\cdot 5x^{\color{blue}{1}}\color{black}{}y^2z^3&\text{Multiply the coefficients and add exponents with same bases} \\ 2\cdot 5\cdot x^{3+1}\cdot y^{5+2}\cdot z^{1+3}&\text{Add exponents and multiply the coefficients} \\ 10x^4y^7z^4&\text{Simplified expression}\end{array}\nonumber\]
Quotient Rule of Exponents
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’s think about Example 6.1.4 . This method of expanding seems to be fine when there are smaller exponents, but what if we were given something like \(\dfrac{a^{199}}{a^{827}}\)? Are we going to expand \(a\) over a thousand times? No way! We need a more sophisticated way in dividing expressions with exponents. Hence, taking a look at Example 6.1.4 , we can see the result is \(a^3\). Notice we could have obtained this answer without expanding but by simply subtracting the exponents:
\[\dfrac{a^5}{a^2}=a^{5-2}=a^3\nonumber\]
This is called the quotient rule of exponents.
Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[\dfrac{a^n}{a^m}=a^{n-m}\nonumber\]
In order to subtract exponents, the bases of the dividend and divisor are required to be the same. Be sure that the denominator exponent is subtracted from the numerator exponent.
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.
Power Rule of Exponents
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’s think about Example 6.1.7 . This method of expanding seems to be fine when there are smaller exponents, but what if we were given something like \((a^{760})^{34}\)? Are we going to expand \(a\) over a twenty-thousand times? No way! We need a more sophisticated way in simplifying expressions with exponents raised to exponents. Hence, taking a look at Example 6.1.7 , we can see the result is \(a^6\). Notice we could have obtained this answer without expanding but by simply multiplying the exponents:
\[(a^2)^3=a^{2\cdot 3}=a^6\nonumber\]
This is called the power rule of exponents.
Let \(a\) be a positive real number and \(n\) and \(m\) be any real number. Then \[(a^n)^m=a^{m\cdot n}\nonumber\]
Furthermore, we can extend the power rule for when we have more than one factor in the base.
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’s think about Example 6.1.8 . This method of expanding seems to be fine when there are smaller exponents, but what if we were given something like \((ab)^{2049}\)? Are we going to expand \(a\) and \(b\) over a two-thousand times? No way! We need a more sophisticated way in simplifying expressions with exponents raised to exponents with more than one factor in the base. Hence, taking a look at Example 6.1.8 , we can see the result is \(a^3 b^3\). Notice we could have obtained this answer without expanding but by simply applying the exponent to each factor in the base:
\[(ab)^3=a^3\cdot b^3=a^3b^3\nonumber\]
This is called the power of a product rule (POP).
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 \(b\).
\[\begin{array}{rl}\left(\dfrac{a}{b}\right)^3&\text{Expand} \\ \left(\dfrac{a}{b}\right)\left(\dfrac{a}{b}\right)\left(\dfrac{a}{b}\right)&\text{Multiply fractions} \\ \dfrac{a^3}{b^3}&\text{Simplified expression}\end{array}\nonumber\]
Notice, this is similar to the POP rule and we can apply the exponent to each numerator and denominator.
Let \(a\) and \(b\) be a positive real numbers and \(n\) be any real number. Then \[\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\nonumber\]
Let’s look at an example where we have to combine all these exponent rules.
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. \(\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\)
Zero as an Exponent
Here we discuss zero as an exponent. This is one of two cases where the exponent isn’t positive. The other case is where the exponents are negative, but we will save that for the next section. Let’s look an example:
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:
\[\begin{array}{rl}(3x^2)^0&\text{Zero power rule} \\ 1&\text{Simplified expression}\end{array}\nonumber\]
Negative Exponents
Another property we consider is expressions with negative exponents.
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 \(a^{-2}=\dfrac{1}{a^2}\).
This example illustrates an important property of exponents. Negative exponents yield the reciprocal of the base. Once we take the reciprocal, the exponent is now positive.
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. \(\left(\dfrac{a}{b}\right)^{-n}=\left(\dfrac{b}{a}\right)^n\)
Negative exponents are combined in several different ways. As a general rule, in a fraction, a base with a negative exponent moves to the other side of the fraction bar as the exponent changes sign.
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.
Properties of Exponents
Putting all the rules together, we can simplify more complex expression containing exponents. Here we apply all the rules of exponents to simplify expressions.
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.
\[\begin{array}{rl} \left(\dfrac{3x^{-2}y^5z^3\cdot 6x^{-6}y^{-2}z^{-3}}{9(x^2y^{-2})^{-3}}\right)^{-3}&\text{Simplify each numeratr and denominator} \\ \left(\dfrac{18x^{-8}y^3z^0}{9x^{-6}y^6}\right)^{-3}&\text{Apply the quotient rule} \\ (2x^{-2}y^{-3}z^0)^{-3}&\text{Apply the POP rule} \\ 2^{-3}x^{-6}y^9z^0&\text{Rewrite only with positive exponents} \\ \dfrac{x^6y^9}{2^3}&\text{Simplify} \\ \dfrac{x^6y^9}{8}&\text{Simplified expression}\end{array}\nonumber\]
Exponent Rules and Properties Homework
Simplify. Be sure to follow the simplifying rules and write answers with positive exponents.
\(4\cdot 4^4\cdot 4^4\)
\(4\cdot 2^2\)
\(3m\cdot 3mn\)
\(2m^4n^2\cdot 4nm^2\)
\((3^3)^4\)
\((4^4)^2\)
\((2u^3v^2)^2\)
\((2a^4)^4\)
\(\dfrac{4^5}{4^3}\)
\(\dfrac{3^2}{3}\)
\(\dfrac{3nm^2}{3n}\)
\(\dfrac{4x^3y^4}{3xy^3}\)
\((x^3y^4\cdot 2x^2y^3)^2\)
\(2x(x^4y^4)^4\)
\(\dfrac{2x^7y^5}{3x^3y\cdot 4x^2y^3}\)
\(\left(\dfrac{(2x)^3}{x^3}\right)^2\)
\(\left(\dfrac{2y^{17}}{(2x^2y^4)^4}\right)^3\)
\(\left(\dfrac{2mn^4\cdot 2m^4n^4}{mn^4}\right)^3\)
\(\dfrac{2xy^5\cdot 2x^2y^3}{2xy^4\cdot y^3}\)
\(\dfrac{q^3r^2\cdot (2p^2q^2r^3)^2}{2p^3}\)
\(\left(\dfrac{zy^3\cdot z^3x^4y^4}{x^3y^3z^3}\right)^4\)
\(\dfrac{2x^2y^2z^6\cdot 2zx^2y^2}{(x^2z^3)^2}\)
\(4\cdot 4^4\cdot 4^2\)
\(3\cdot 3^3\cdot 3^2\)
\(3x\cdot 4x^2\)
\(x^2y^4\cdot xy^2\)
\((4^3)^4\)
\((3^2)^3\)
\((xy)^3\)
\((2xy)^4\)
\(\dfrac{3^7}{3^3}\)
\(\dfrac{3^4}{3}\)
\(\dfrac{x^2y^4}{4xy}\)
\(\dfrac{xy^3}{4xy}\)
\((u^2v^2\cdot 2u^4)^3\)
\(\dfrac{3vu^5\cdot 2v^3}{uv^2\cdot 2u^3v}\)
\(\dfrac{2ba^7\cdot 2b^4}{ba^2\cdot 3a^3b^4}\)
\(\dfrac{2a^2b^2a^7}{(ba^4)^2}\)
\(\dfrac{yx^2\cdot (y^4)^2}{2y^4}\)
\(\dfrac{n^3(n^4)^2}{2mn}\)
\(\dfrac{(2y^3x^2)^2}{2x^2y^4\cdot x^2}\)
\(\dfrac{2x^4y^5\cdot 2z^{10}x^2y^7}{(xy^2z^2)^4}\)
\(\left(\dfrac{2q^3p^3r^4\cdot 2p^3}{(qrp^3)^2}\right)^4\)
\(2x^4y^{-2}\cdot (2xy^3)^4\)
\((a^4b^{-3})^3\cdot 2a^3b^{-2}\)
\((2x^2y^2)^4x^{-4}\)
\((x^3y^4)^3\cdot x^{-4}y^4\)
\(\dfrac{2x^{-3}y^2}{3x^{-3}y^3\cdot 3x^0}\)
\(\dfrac{4xy^{-3}\cdot x^{-4}y^0}{4y^{-1}}\)
\(\dfrac{u^2v^{-1}}{2u^0v^4\cdot 2uv}\)
\(\dfrac{u^2}{4u^0v^3\cdot 3v^2}\)
\(\dfrac{2y}{(x^0y^2)^4}\)
\(\left(\dfrac{2a^2b^3}{a^{-1}}\right)^4\)
\(\dfrac{2nm^4}{(2m^2n^2)^4}\)
\(\dfrac{(2mn)^4}{m^0n^{-2}}\)
\(\dfrac{y^3\cdot x^{-3}y^2}{(x^4y^2)^3}\)
\(\dfrac{2u^{-2}v^3\cdot (2uv^4)^{-1}}{2u^{-4}v^0}\)
\(\left(\dfrac{2x^0\cdot y^4}{y^4}\right)^3\)
\(\dfrac{y(2x^4y^2)^2}{2x^4y^0}\)
\(\dfrac{2yzx^2}{2x^4y^4z^{-2}\cdot (zy^2)^4}\)
\(\dfrac{2kh^0\cdot 2h^{-3}k^0}{(2kj^3)^2}\)
\(\dfrac{(cb^3)^2\cdot 2a^{-3}b^2}{(a^3b^{-2}c^3)^3}\)
\(\dfrac{(yx^{-4}z^2)^{-1}}{z^3\cdot x^2y^3z^{-1}}\)
\(2a^{-2}b^{-3}\cdot (2a^0b^4)^4\)
\(2x^3y^2\cdot (2x^3)^0\)
\((m^0n^3\cdot 2m^{-3}n^{-3})^0\)
\(2m^{-1}n^{-3}\cdot (2m^{-1}n^{-3})^4\)
\(\dfrac{3y^3}{3yx^3\cdot 2x^4y^{-3}}\)
\(\dfrac{3x^3y^2}{4y^{-2}\cdot 3x^{-2}y^{-4}}\)
\(\dfrac{2xy^2\cdot 4x^3y^{-4}}{4x^{-4}y^{-4}\cdot 4x}\)
\(\dfrac{2x^{-2}y^2}{4yx^2}\)
\(\dfrac{(a^4)^4}{2b}\)
\(\left(\dfrac{2y^{-4}}{x^2}\right)^{-2}\)
\(\dfrac{2y^2}{(x^4y^0)^{-4}}\)
\(\dfrac{2x^{-3}}{(x^4y^{-3})^{-1}}\)
\(\dfrac{2x^{-2}y^0\cdot 2xy^4}{(xy^0)^{-1}}\)
\(\dfrac{2yx^2\cdot x^{-2}}{(2x^0y^4)^{-1}}\)
\(\dfrac{u^{-3}v^{-4}}{2v(2u^{-3}v^4)^0}\)
\(\dfrac{b^{-1}}{(2a^4b^0)^0\cdot 2a^{-3}b^2}\)
\(\dfrac{2b^4c^{-2}\cdot (2b^3c^2)^{-4}}{a^{-2}b^4}\)
\(\left(\dfrac{(2x^{-3}y^0z^{-1})^3\cdot x^{-3}y^2}{2x^3}\right)^{-2}\)
\(\dfrac{2q^4\cdot m^2p^2q^4}{(2m^{-4}p^2)^3}\)
\(\dfrac{2mpn^{-3}}{(m^0n^{-4}p^2)^3\cdot 2n^2p^0}\)