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Guias de estudo > ALGEBRA / TRIG I

Simplifying Complex Expressions II

Learning Outcomes

  • Simplify complex expressions using a combination of exponent rules
  • Simplify quotients that require a combination of the properties of exponents

Simplify expressions using a combination of exponent rules

Once the rules of exponents are understood, you can begin simplifying more complicated expressions. There are many applications and formulas that make use of exponents, and sometimes expressions can get pretty cluttered.  We already looked at how to simplify exponential expressions in the previous section, but we are now going to show some more complex examples of exponential expressions and look at some strategies for how we can successfully use the properties of exponents to simplify these expressions. Simplifying an expression before evaluating can often make the computation easier, as you will see in the following example which makes use of the quotient rule to simplify before substituting 44 for xx.

Example

Evaluate 24x82x5 \displaystyle \frac{24{{x}^{8}}}{2{{x}^{5}}} when x=4x=4.

Answer: Separate into numerical and variable factors. (242)(x8x5) \displaystyle \left( \frac{24}{2} \right)\left( \frac{{{x}^{8}}}{{{x}^{5}}} \right) Divide coefficients, and subtract the exponents of the variables. 12(x8−5) \displaystyle 12\left( {{x}^{8-5}} \right) Simplify. 12x3 \displaystyle 12{{x}^{3}} Substitute the value 44 for the variable xx. (12)(43)=12⋅64 \displaystyle (12)({{4}^{3}})=12\cdot 64

Answer

24x82x5[/latex]=[latex]768 \displaystyle \frac{24{{x}^{8}}}{2{{x}^{5}}}[/latex] = [latex]768

Example

Evaluate 24x8y2(2x3y)2 \displaystyle \frac{24{{x}^{8}}{{y}^{2}}}{{{(2{{x}^{3}}y)}^{2}}} when x=4x=4 and y=−2y=-2.

Answer: In the denominator, notice that a product is being raised to a power.   Use the rules of exponents to simplify the denominator.

(2x3y)2=22x3⋅2y2=22x6y2=4x6y2 \displaystyle{\left(2{x}^{3}y\right)}^{2}={2}^{2}{x}^{3\cdot 2}{y}^{2}={2}^{2}{x}^{6}{y}^{2}={4x^{6}y^{2}}

Here is the fraction with a simplified denominator:

24x8y24x6y2\frac{24x^{8}y^{2}}{4x^{6}y^{2}}

Separate into numerical and variable factors to simplify further.

(244)(x8x6)(y2y2) \displaystyle \left( \frac{24}{4} \right)\left( \frac{{{x}^{8}}}{{{x}^{6}}} \right)\left( \frac{{{y}^{2}}}{{{y}^{2}}}\right)

Divide coefficients, use the Quotient Rule to divide the variables—subtract the exponents.

6(x8−6)(y2−2) \displaystyle 6\left( {{x}^{8-6}} \right)\left( {{y}^{2-2}} \right)

Simplify. Remember that y0y^{0} is 11.

6x2y0=6x2 \displaystyle 6{{x}^{2}}{{y}^{0}}=6{{x}^{2}}

Substitute the value 44 for the variable xx.

(6)(42)=6⋅16 \displaystyle (6)({{4}^{2}})=6\cdot 16

Answer

24x8y2(2x3y)2=96[/latex]when[latex]x=4[/latex] and[latex]y=−2 \displaystyle \frac{24{{x}^{8}}{{y}^{2}}}{{{(2{{x}^{3}}y)}^{2}}}=96[/latex] when [latex]x=4[/latex] and [latex]y=-2

Try It

[ohm_question]2856[/ohm_question]
Notice that you could have worked this problem by substituting 44 for xx and 22 for yy in the original expression. You would still get the answer of 9696, but the computation would be much more complex. Notice that you didn’t even need to use the value of yy to evaluate the above expression. In the following video you are shown examples of evaluating an exponential expression for given numbers. https://youtu.be/mD06EyGja2w Usually, it is easier to simplify the expression before substituting any values for your variables, but you will get the same answer either way. In the next examples, you will see how to simplify expressions using different combinations of the rules for exponents.

Example

Simplify. a2(a5)3a^{2}\left(a^{5}\right)^{3}

Answer: Raise a5a^{5} to the power of 33 by multiplying the exponents together (the Power Rule).

a2a5⋅3 \displaystyle {{a}^{2}}{{a}^{5\cdot 3}}

Since the exponents share the same base, a, they can be combined (the Product Rule).

a2a15a2+15 \displaystyle {{a}^{2}}{{a}^{15}}\\{{a}^{2+15}}

Answer

a2(a5)3=a17 \displaystyle {{a}^{2}}{{({{a}^{5}})}^{3}}={{a}^{17}}

Try It

[ohm_question]15516[/ohm_question]
The following examples require the use of all the exponent rules we have learned so far. Remember that the product, power, and quotient rules apply when your terms have the same base.

Example

Simplify. a2(a5)38a8 \displaystyle \frac{{{a}^{2}}{{({{a}^{5}})}^{3}}}{8{{a}^{8}}}

Answer: Use the order of operations. Parentheses, Exponents, Multiply/ Divide, Add/ Subtract There is nothing inside parentheses or brackets that we can simplify further, so we will evaluate exponents first. Use the Power Rule to simplify (a5)3\left(a^{5}\right)^{3}.

(a5)3=a5⋅3=a15\left(a^{5}\right)^{3}=a^{5\cdot{3}}=a^{15}

The expression now looks like this:

a2a158a8 \displaystyle\frac{{{a}^{2}}{{a}^{15}}}{8{{a}^{8}}}

Now we can multiply, using the Product Rule to simplify the numerator because the bases are the same.

a2a15=a17 \displaystyle{{a}^{2}}{{a}^{15}}=a^{17}, and the expression looks like this:

a178a8\displaystyle\frac{{{{a}^{17}}}}{{8{{a}^{8}}}}

Now we can divide using the Quotient Rule.

a17−88 \displaystyle \frac{{{a}^{17-8}}}{8}

Answer

a2(a5)38a8=a98 \displaystyle \frac{{{a}^{2}}{{({{a}^{5}})}^{3}}}{8{{a}^{8}}}=\frac{{{a}^{9}}}{8}

Try It

[ohm_question]14055[/ohm_question]

Simplify Expressions With Negative Exponents

Now we will add the last layer to our exponent simplifying skills and practice simplifying compound expressions that have negative exponents in them. It is standard convention to write exponents as positive because it is easier for the user to understand the value associated with positive exponents, rather than negative exponents. Use the following summary of negative exponents to help you simplify expressions with negative exponents.

Rules for Negative Exponents

With a, b, m, and n not equal to zero, and m and n as integers, the following rules apply:

a−m=1ama^{-m}=\frac{1}{a^{m}}

1a−m=am\frac{1}{a^{-m}}=a^{m}

a−nb−m=bman\frac{a^{-n}}{b^{-m}}=\frac{b^m}{a^n}

When you are simplifying expressions that have many layers of exponents, it is often hard to know where to start. It is common to start in one of two ways:
  • Rewrite negative exponents as positive exponents
  • Apply the product rule to eliminate any "outer" layer exponents such as in the following term: (5y3)2\left(5y^3\right)^2
We will explore this idea with the following example: Simplify. (4x3)5⋅  (2x2)−4 \displaystyle {{\left( 4{{x}^{3}} \right)}^{5}}\cdot \,\,{{\left( 2{{x}^{2}} \right)}^{-4}} Write your answer with positive exponents. The table below shows how to simplify the same expression in two different ways, rewriting negative exponents as positive first, and applying the product rule for exponents first. You will see that there is a column for each method that describes the exponent rule or other steps taken to simplify the expression.
Rewrite with positive Exponents First Description of Steps Taken Apply the Product Rule for Exponents First Description of Steps Taken
(4x3)5(2x2)4 \frac{\left(4x^{3}\right)^{5}}{\left(2x^{2}\right)^{4}} move the term (2x2)−4{{\left( 2{{x}^{2}} \right)}^{-4}} to the denominator with a positive exponent (45x15)(2−4x−8) \left(4^5x^{15}\right)\left(2^{-4}x^{-8}\right)  Apply the exponent of 5 to each term in expression on the left, and the exponent of -4 to each term in the expression on the right.
 (45x15)(24x8)\frac{\left(4^5x^{15}\right)}{\left(2^4x^{8}\right)} Use the product rule to apply the outer exponents to the terms inside each set of parentheses. (45)(2−4)(x15⋅x−8)\left(4^5\right)\left(2^{-4}\right)\left(x^{15}\cdot{x^{-8}}\right) Regroup the numerical terms and the variables to make combining like terms easier
 (4524)(x15x8)\left(\frac{4^5}{2^4}\right)\left(\frac{x^{15}}{x^{8}}\right) Regroup the numerical terms and the variables to make combining like terms easier (45)(2−4)(x15−8)\left(4^5\right)\left(2^{-4}\right)\left(x^{15-8}\right)  Use the rule for multiplying terms with exponents to simplify the x terms
 (4524)(x15−8)\left(\frac{4^5}{2^4}\right)\left(x^{15-8}\right) Use the quotient rule to simplify the x terms (4524)(x7)\left(\frac{4^5}{2^4}\right)\left(x^{7}\right)  Rewrite all the negative exponents with positive exponents
 (1,02416)(x7)\left(\frac{1,024}{16}\right)\left(x^{7}\right) Expand the numerical terms (1,02416)(x7)\left(\frac{1,024}{16}\right)\left(x^{7}\right)  Expand the numerical terms
  64x764x^{7} Divide the numerical terms  64x764x^{7}  Divide the numerical terms
If you compare the two columns that describe the steps that were taken to simplify the expression, you will see that they are all nearly the same, except the order is changed slightly. Neither way is better or more correct than the other, it truly is a matter of preference.

Example

Simplify (t3)2(t2)−8\frac{\left(t^{3}\right)^2}{\left(t^2\right)^{-8}} Write your answer with positive exponents.

Answer: We can either rewrite this expression with positive exponents first or use the Product Raised to a Power Rule first. Let's start by simplifying the numerator and denominator using the Product Raised to a Power Rule. Numerator: (t3)2=t3⋅2=t6\left(t^{3}\right)^2=t^{3\cdot{2}}=t^6 Denominator: (t2)−8=t2⋅−8=t−16\left(t^2\right)^{-8}=t^{2\cdot{-8}}=t^{-16} Now the expression looks like this:

t6t−16\frac{t^6}{t^{-16}}

We can use the quotient rule because we have the same base. Quotient Rule: t6t−16=t6−(−16)=t6+16=t22\frac{t^6}{t^{-16}}=t^{6-\left(-16\right)}=t^{6+16}=t^{22}

Answer

(t3)2(t2)−8=t22\frac{\left(t^{3}\right)^2}{\left(t^2\right)^{-8}}=t^{22}

Try It

[ohm_question]133473[/ohm_question]

Example

Simplify (5x)−2yx3y−1\frac{\left(5x\right)^{-2}y}{x^3y^{-1}} Write your answer with positive exponents.

Answer: This time, let's start by rewriting the terms in the expression so they have positive exponents. The terms with negative exponents in the top will go to the bottom of the fraction, and the terms with negative exponents in the bottom will go to the top.

(5x)−2yx3y−1 =(y1)yx3(5x)2\begin{array}{c}\frac{\left(5x\right)^{-2}y}{x^3y^{-1}}\\\text{ }\\=\frac{\left({y^{1}}\right)y}{x^3\left(5x\right)^{2}}\end{array}

  Note how we left the single yy term in the top because it did not have a negative exponent on it, and we left the x3x^3 term in the bottom because it did not have a negative exponent on it.   Now we can apply the Product Raised to a Power Rule:

yy152x3x2\frac{yy^{1}}{5^{2}x^3x^{2}}

Use the product rule to simplify further:

yy152x3x2=y225x3+2=y225x5\frac{yy^{1}}{5^{2}x^3x^{2}}=\frac{y^2}{25x^{3+2}}=\frac{y^2}{25x^5}

We can't simplify any further, so our answer is

Answer

(5x)−2yx3y−1=y225x5\frac{\left(5x\right)^{-2}y}{x^3y^{-1}}=\frac{y^2}{25x^5}

In the next section, you will learn how to write very large and very small numbers using exponents. This practice is widely used in science and engineering.  

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