Whether you are a student working on math assignments or a professional dealing with equations as part of your job, learning to simplify expressions is a valuable investment in your mathematical education and career. The exponent rules chart that can be used for simplifying algebraic expressions is given below: To simplify this expression, let us first open the bracket by multiplying 4b to both the terms written inside. [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}=\frac{{f}^{14}}{{e}^{14}}[/latex], [latex]\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex], [latex]\begin{array}{ccc}\hfill {\left({e}^{-2}{f}^{2}\right)}^{7}& =& {\left(\frac{{f}^{2}}{{e}^{2}}\right)}^{7}\hfill \\ & =& \frac{{\left({f}^{2}\right)}^{7}}{{\left({e}^{2}\right)}^{7}}\hfill \\ & =& \frac{{f}^{2\cdot 7}}{{e}^{2\cdot 7}}\hfill \\ & =& \frac{{f}^{14}}{{e}^{14}}\hfill \end{array}[/latex], [latex]{\left(\frac{a}{b}\right)}^{n}=\frac{{a}^{n}}{{b}^{n}}[/latex], CC licensed content, Specific attribution, http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@5.2, http://cnx.org/contents/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1/Preface, [latex]\left(3a\right)^{7}\cdot\left(3a\right)^{10} [/latex], [latex]\left(\left(3a\right)^{7}\right)^{10} [/latex], [latex]\left(3a\right)^{7\cdot10} [/latex], [latex]{\left(a\cdot b\right)}^{n}={a}^{n}\cdot {b}^{n}[/latex], [latex]\left(-3\right)^{5}\cdot \left(-3\right)[/latex], [latex]{x}^{2}\cdot {x}^{5}\cdot {x}^{3}[/latex], [latex]{t}^{5}\cdot {t}^{3}={t}^{5+3}={t}^{8}[/latex], [latex]{\left(-3\right)}^{5}\cdot \left(-3\right)={\left(-3\right)}^{5}\cdot {\left(-3\right)}^{1}={\left(-3\right)}^{5+1}={\left(-3\right)}^{6}[/latex], [latex]{\left(\frac{2}{y}\right)}^{4}\cdot \left(\frac{2}{y}\right)[/latex], [latex]{t}^{3}\cdot {t}^{6}\cdot {t}^{5}[/latex], [latex]{\left(\frac{2}{y}\right)}^{5}[/latex], [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}[/latex], [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}[/latex], [latex]\frac{{\left(-2\right)}^{14}}{{\left(-2\right)}^{9}}={\left(-2\right)}^{14 - 9}={\left(-2\right)}^{5}[/latex], [latex]\frac{{t}^{23}}{{t}^{15}}={t}^{23 - 15}={t}^{8}[/latex], [latex]\frac{{\left(z\sqrt{2}\right)}^{5}}{z\sqrt{2}}={\left(z\sqrt{2}\right)}^{5 - 1}={\left(z\sqrt{2}\right)}^{4}[/latex], [latex]\frac{{\left(-3\right)}^{6}}{-3}[/latex], [latex]\frac{{\left(e{f}^{2}\right)}^{5}}{{\left(e{f}^{2}\right)}^{3}}[/latex], [latex]{\left(e{f}^{2}\right)}^{2}[/latex], [latex]{\left({x}^{2}\right)}^{7}[/latex], [latex]{\left({\left(2t\right)}^{5}\right)}^{3}[/latex], [latex]{\left({\left(-3\right)}^{5}\right)}^{11}[/latex], [latex]{\left({x}^{2}\right)}^{7}={x}^{2\cdot 7}={x}^{14}[/latex], [latex]{\left({\left(2t\right)}^{5}\right)}^{3}={\left(2t\right)}^{5\cdot 3}={\left(2t\right)}^{15}[/latex], [latex]{\left({\left(-3\right)}^{5}\right)}^{11}={\left(-3\right)}^{5\cdot 11}={\left(-3\right)}^{55}[/latex], [latex]{\left({\left(3y\right)}^{8}\right)}^{3}[/latex], [latex]{\left({t}^{5}\right)}^{7}[/latex], [latex]{\left({\left(-g\right)}^{4}\right)}^{4}[/latex], [latex]\frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}[/latex], [latex]\frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}[/latex], [latex]\begin{array}\text{ }\frac{c^{3}}{c^{3}} \hfill& =c^{3-3} \\ \hfill& =c^{0} \\ \hfill& =1\end{array}[/latex], [latex]\begin{array}{ccc}\hfill \frac{-3{x}^{5}}{{x}^{5}}& =& -3\cdot \frac{{x}^{5}}{{x}^{5}}\hfill \\ & =& -3\cdot {x}^{5 - 5}\hfill \\ & =& -3\cdot {x}^{0}\hfill \\ & =& -3\cdot 1\hfill \\ & =& -3\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill \frac{{\left({j}^{2}k\right)}^{4}}{\left({j}^{2}k\right)\cdot {\left({j}^{2}k\right)}^{3}}& =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{1+3}}\hfill & \text{Use the product rule in the denominator}.\hfill \\ & =& \frac{{\left({j}^{2}k\right)}^{4}}{{\left({j}^{2}k\right)}^{4}}\hfill & \text{Simplify}.\hfill \\ & =& {\left({j}^{2}k\right)}^{4 - 4}\hfill & \text{Use the quotient rule}.\hfill \\ & =& {\left({j}^{2}k\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 1& \end{array}[/latex], [latex]\begin{array}{cccc}\hfill \frac{5{\left(r{s}^{2}\right)}^{2}}{{\left(r{s}^{2}\right)}^{2}}& =& 5{\left(r{s}^{2}\right)}^{2 - 2}\hfill & \text{Use the quotient rule}.\hfill \\ & =& 5{\left(r{s}^{2}\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 5\cdot 1\hfill & \text{Use the zero exponent rule}.\hfill \\ & =& 5\hfill & \text{Simplify}.\hfill \end{array}[/latex], [latex]\frac{{\left(d{e}^{2}\right)}^{11}}{2{\left(d{e}^{2}\right)}^{11}}[/latex], [latex]\frac{{w}^{4}\cdot {w}^{2}}{{w}^{6}}[/latex], [latex]\frac{{t}^{3}\cdot {t}^{4}}{{t}^{2}\cdot {t}^{5}}[/latex], [latex]\frac{{\theta }^{3}}{{\theta }^{10}}[/latex], [latex]\frac{{z}^{2}\cdot z}{{z}^{4}}[/latex], [latex]\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}[/latex], [latex]\frac{{\theta }^{3}}{{\theta }^{10}}={\theta }^{3 - 10}={\theta }^{-7}=\frac{1}{{\theta }^{7}}[/latex], [latex]\frac{{z}^{2}\cdot z}{{z}^{4}}=\frac{{z}^{2+1}}{{z}^{4}}=\frac{{z}^{3}}{{z}^{4}}={z}^{3 - 4}={z}^{-1}=\frac{1}{z}[/latex], [latex]\frac{{\left(-5{t}^{3}\right)}^{4}}{{\left(-5{t}^{3}\right)}^{8}}={\left(-5{t}^{3}\right)}^{4 - 8}={\left(-5{t}^{3}\right)}^{-4}=\frac{1}{{\left(-5{t}^{3}\right)}^{4}}[/latex], [latex]\frac{{\left(-3t\right)}^{2}}{{\left(-3t\right)}^{8}}[/latex], [latex]\frac{{f}^{47}}{{f}^{49}\cdot f}[/latex], [latex]\frac{1}{{\left(-3t\right)}^{6}}[/latex], [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}[/latex], [latex]\frac{-7z}{{\left(-7z\right)}^{5}}[/latex], [latex]{b}^{2}\cdot {b}^{-8}={b}^{2 - 8}={b}^{-6}=\frac{1}{{b}^{6}}[/latex], [latex]{\left(-x\right)}^{5}\cdot {\left(-x\right)}^{-5}={\left(-x\right)}^{5 - 5}={\left(-x\right)}^{0}=1[/latex], [latex]\frac{-7z}{{\left(-7z\right)}^{5}}=\frac{{\left(-7z\right)}^{1}}{{\left(-7z\right)}^{5}}={\left(-7z\right)}^{1 - 5}={\left(-7z\right)}^{-4}=\frac{1}{{\left(-7z\right)}^{4}}[/latex], [latex]\frac{{25}^{12}}{{25}^{13}}[/latex], [latex]{t}^{-5}=\frac{1}{{t}^{5}}[/latex], [latex]{\left(a{b}^{2}\right)}^{3}[/latex], [latex]{\left(-2{w}^{3}\right)}^{3}[/latex], [latex]\frac{1}{{\left(-7z\right)}^{4}}[/latex], [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}[/latex], [latex]{\left(a{b}^{2}\right)}^{3}={\left(a\right)}^{3}\cdot {\left({b}^{2}\right)}^{3}={a}^{1\cdot 3}\cdot {b}^{2\cdot 3}={a}^{3}{b}^{6}[/latex], [latex]2{t}^{15}={\left(2\right)}^{15}\cdot {\left(t\right)}^{15}={2}^{15}{t}^{15}=32,768{t}^{15}[/latex], [latex]{\left(-2{w}^{3}\right)}^{3}={\left(-2\right)}^{3}\cdot {\left({w}^{3}\right)}^{3}=-8\cdot {w}^{3\cdot 3}=-8{w}^{9}[/latex], [latex]\frac{1}{{\left(-7z\right)}^{4}}=\frac{1}{{\left(-7\right)}^{4}\cdot {\left(z\right)}^{4}}=\frac{1}{2,401{z}^{4}}[/latex], [latex]{\left({e}^{-2}{f}^{2}\right)}^{7}={\left({e}^{-2}\right)}^{7}\cdot {\left({f}^{2}\right)}^{7}={e}^{-2\cdot 7}\cdot {f}^{2\cdot 7}={e}^{-14}{f}^{14}=\frac{{f}^{14}}{{e}^{14}}[/latex], [latex]{\left({g}^{2}{h}^{3}\right)}^{5}[/latex], [latex]{\left(-3{y}^{5}\right)}^{3}[/latex], [latex]\frac{1}{{\left({a}^{6}{b}^{7}\right)}^{3}}[/latex], [latex]{\left({r}^{3}{s}^{-2}\right)}^{4}[/latex], [latex]\frac{1}{{a}^{18}{b}^{21}}[/latex], [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}[/latex], [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}[/latex], [latex]{\left(\frac{-1}{{t}^{2}}\right)}^{27}[/latex], [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}[/latex], [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}[/latex], [latex]{\left(\frac{4}{{z}^{11}}\right)}^{3}=\frac{{\left(4\right)}^{3}}{{\left({z}^{11}\right)}^{3}}=\frac{64}{{z}^{11\cdot 3}}=\frac{64}{{z}^{33}}[/latex], [latex]{\left(\frac{p}{{q}^{3}}\right)}^{6}=\frac{{\left(p\right)}^{6}}{{\left({q}^{3}\right)}^{6}}=\frac{{p}^{1\cdot 6}}{{q}^{3\cdot 6}}=\frac{{p}^{6}}{{q}^{18}}[/latex], [latex]{\\left(\frac{-1}{{t}^{2}}\\right)}^{27}=\frac{{\\left(-1\\right)}^{27}}{{\\left({t}^{2}\\right)}^{27}}=\frac{-1}{{t}^{2\cdot 27}}=\frac{-1}{{t}^{54}}=-\frac{1}{{t}^{54}}[/latex], [latex]{\left({j}^{3}{k}^{-2}\right)}^{4}={\left(\frac{{j}^{3}}{{k}^{2}}\right)}^{4}=\frac{{\left({j}^{3}\right)}^{4}}{{\left({k}^{2}\right)}^{4}}=\frac{{j}^{3\cdot 4}}{{k}^{2\cdot 4}}=\frac{{j}^{12}}{{k}^{8}}[/latex], [latex]{\left({m}^{-2}{n}^{-2}\right)}^{3}={\left(\frac{1}{{m}^{2}{n}^{2}}\right)}^{3}=\frac{{\left(1\right)}^{3}}{{\left({m}^{2}{n}^{2}\right)}^{3}}=\frac{1}{{\left({m}^{2}\right)}^{3}{\left({n}^{2}\right)}^{3}}=\frac{1}{{m}^{2\cdot 3}\cdot {n}^{2\cdot 3}}=\frac{1}{{m}^{6}{n}^{6}}[/latex], [latex]{\left(\frac{{b}^{5}}{c}\right)}^{3}[/latex], [latex]{\left(\frac{5}{{u}^{8}}\right)}^{4}[/latex], [latex]{\left(\frac{-1}{{w}^{3}}\right)}^{35}[/latex], [latex]{\left({p}^{-4}{q}^{3}\right)}^{8}[/latex], [latex]{\left({c}^{-5}{d}^{-3}\right)}^{4}[/latex], [latex]\frac{1}{{c}^{20}{d}^{12}}[/latex], [latex]{\left(6{m}^{2}{n}^{-1}\right)}^{3}[/latex], [latex]{17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}[/latex], [latex]{\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}[/latex], [latex]\left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)[/latex], [latex]{\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}[/latex], [latex]\frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}[/latex], [latex]\begin{array}{cccc}\hfill {\left(6{m}^{2}{n}^{-1}\right)}^{3}& =& {\left(6\right)}^{3}{\left({m}^{2}\right)}^{3}{\left({n}^{-1}\right)}^{3}\hfill & \text{The power of a product rule}\hfill \\ & =& {6}^{3}{m}^{2\cdot 3}{n}^{-1\cdot 3}\hfill & \text{The power rule}\hfill \\ & =& \text{ }216{m}^{6}{n}^{-3}\hfill & \text{Simplify}.\hfill \\ & =& \frac{216{m}^{6}}{{n}^{3}}\hfill & \text{The negative exponent rule}\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill {17}^{5}\cdot {17}^{-4}\cdot {17}^{-3}& =& {17}^{5 - 4-3}\hfill & \text{The product rule}\hfill \\ & =& {17}^{-2}\hfill & \text{Simplify}.\hfill \\ & =& \frac{1}{{17}^{2}}\text{ or }\frac{1}{289}\hfill & \text{The negative exponent rule}\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill {\left(\frac{{u}^{-1}v}{{v}^{-1}}\right)}^{2}& =& \frac{{\left({u}^{-1}v\right)}^{2}}{{\left({v}^{-1}\right)}^{2}}\hfill & \text{The power of a quotient rule}\hfill \\ & =& \frac{{u}^{-2}{v}^{2}}{{v}^{-2}}\hfill & \text{The power of a product rule}\hfill \\ & =& {u}^{-2}{v}^{2-\left(-2\right)}& \text{The quotient rule}\hfill \\ & =& {u}^{-2}{v}^{4}\hfill & \text{Simplify}.\hfill \\ & =& \frac{{v}^{4}}{{u}^{2}}\hfill & \text{The negative exponent rule}\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill \left(-2{a}^{3}{b}^{-1}\right)\left(5{a}^{-2}{b}^{2}\right)& =& -2\cdot 5\cdot {a}^{3}\cdot {a}^{-2}\cdot {b}^{-1}\cdot {b}^{2}\hfill & \text{Commutative and associative laws of multiplication}\hfill \\ & =& -10\cdot {a}^{3 - 2}\cdot {b}^{-1+2}\hfill & \text{The product rule}\hfill \\ & =& -10ab\hfill & \text{Simplify}.\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill {\left({x}^{2}\sqrt{2}\right)}^{4}{\left({x}^{2}\sqrt{2}\right)}^{-4}& =& {\left({x}^{2}\sqrt{2}\right)}^{4 - 4}\hfill & \text{The product rule}\hfill \\ & =& \text{ }{\left({x}^{2}\sqrt{2}\right)}^{0}\hfill & \text{Simplify}.\hfill \\ & =& 1\hfill & \text{The zero exponent rule}\hfill \end{array}[/latex], [latex]\begin{array}{cccc}\hfill \frac{{\left(3{w}^{2}\right)}^{5}}{{\left(6{w}^{-2}\right)}^{2}}& =& \frac{{\left(3\right)}^{5}\cdot {\left({w}^{2}\right)}^{5}}{{\left(6\right)}^{2}\cdot {\left({w}^{-2}\right)}^{2}}\hfill & \text{The power of a product rule}\hfill \\ & =& \frac{{3}^{5}{w}^{2\cdot 5}}{{6}^{2}{w}^{-2\cdot 2}}\hfill & \text{The power rule}\hfill \\ & =& \frac{243{w}^{10}}{36{w}^{-4}}\hfill & \text{Simplify}.\hfill \\ & =& \frac{27{w}^{10-\left(-4\right)}}{4}\hfill & \text{The quotient rule and reduce fraction}\hfill \\ & =& \frac{27{w}^{14}}{4}\hfill & \text{Simplify}.\hfill \end{array}[/latex], [latex]{\left(2u{v}^{-2}\right)}^{-3}[/latex], [latex]{x}^{8}\cdot {x}^{-12}\cdot x[/latex], [latex]{\left(\frac{{e}^{2}{f}^{-3}}{{f}^{-1}}\right)}^{2}[/latex], [latex]\left(9{r}^{-5}{s}^{3}\right)\left(3{r}^{6}{s}^{-4}\right)[/latex], [latex]{\left(\frac{4}{9}t{w}^{-2}\right)}^{-3}{\left(\frac{4}{9}t{w}^{-2}\right)}^{3}[/latex], [latex]\frac{{\left(2{h}^{2}k\right)}^{4}}{{\left(7{h}^{-1}{k}^{2}\right)}^{2}}[/latex]. It does not intend to find the value of an unknown quantity. Give it a try now and see how it can simplify your algebraic expressions and make your math problems a breeze! (10^5=) The calculator should display the number 100,000, because that's equal to 10 5. When we use rational exponents, we can apply the properties of exponents to simplify expressions. [latex]\begin{array}{ccc}\hfill {\left({x}^{2}\right)}^{3}& =& \stackrel{{3\text{ factors}}}{{{\left({x}^{2}\right)\cdot \left({x}^{2}\right)\cdot \left({x}^{2}\right)}}}\hfill \\ & =& \hfill \stackrel{{3\text{ factors}}}{{{\left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)\cdot \left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)\cdot \left(\stackrel{{2\text{ factors}}}{{\overbrace{x\cdot x}}}\right)}}}\\ & =& x\cdot x\cdot x\cdot x\cdot x\cdot x\hfill \\ & =& {x}^{6}\hfill \end{array}[/latex], [latex]{\left({a}^{m}\right)}^{n}={a}^{m\cdot n}[/latex]. Write answers with positive exponents. Simplify, Simplify (a12b)12(ab12) 1 comment ( 7 votes) Upvote Downvote Flag more htom 2 years ago well what if something was like 1/2 to the power of 7 how would you solve that? Typing Exponents. With a negative exponent, this causes the expression to reciprocate and change exponent to positive, so start with 1/ (4096)^ (5/6) = 1/4^5 = 1/1024. This website uses cookies to ensure you get the best experience on our website. She holds a master's degree in Learning and Technology. There are several steps you can follow to simplify an algebraic expression: Combine like terms: The first step in simplifying an expression is to look for terms with the same variables and exponents and combine them using the appropriate operations. So, adding these two pairs of like terms will result in (6x - 3x) + (-x2 + x2). For example, 1/2 (x + 4) can be simplified as x/2 + 2. Step 1, how do i find my safe credit union account number, how to write a number in expanded form in two ways, simplify expressions with rational exponents calculator. simplify, solve for, expand, factor, rationalize. The expression inside the parentheses is multiplied twice because it has an exponent of 2. Can we simplify the result? Answer Comment ( 3 votes) Upvote Downvote Flag more It helped me pass my exam and the test questions are very similar to the practice quizzes on Study.com. Exponent rules can be used to simplify terms with exponents. To simplify this expression, we need to use the concept of multiplication of algebraic expressions. Some useful properties include. Here is an example: 2x^2+x (4x+3) We strive to deliver products of the highest quality to our customers. I would definitely recommend Study.com to my colleagues. Remove unnecessary terms: If a term has a coefficient of 0, it can be removed from the expression since it has no effect on the value. The simplify calculator will then show you the steps to help you learn how to simplify your algebraic expression on your own. Being a virtual student, it's been able to help study and understand and breakdown concepts that I was not previously aware of. Solution: Given, Daniel bought 5 pencils each for $x. Kathryn teaches college math. With Cuemath, you will learn visually and be surprised by the outcomes. Mathway requires javascript and a modern browser. This will give us x^3-7, which is -4 and y^8-3, which is 5. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number. Do not simplify further. Simplifying dividing algebraic expressions, solve 3x3 systems of linear equations with TI-84 calculator, solving parabola functions, Easiest way to Factor a third-degree polynomial. simplify rational or radical expressions with our free step-by-step math First Law of Exponents If a and b are positive integers and x is a real number. By using the distributive property of simplifying expression, it can be simplified as. Use the product rule to simplify each expression. The E13 portion of the result represents the exponent 13 of ten, so there are a maximum of approximately [latex]1.3\times {10}^{13}[/latex] bits of data in that one-hour film.

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