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QuestionSimplify these expressions with positive indices:
1. (11−18)(11 - 18)
2. (3w6)3(3 w^{6})^{3}
3. (22s5)3(2^{2} s^{5})^{3}
4. (pq5)3\left(\frac{p}{q^{5}}\right)^{3}
5. (−m3n5)5\left(\frac{-m^{3}}{n^{5}}\right)^{5}
6. (s3)−5×s7(s^{3})^{-5} \times s^{7}
7. u−4÷(u5)−2u^{-4} \div (u^{5})^{-2}
8. d3÷(d−5)2d−2\frac{d^{3} \div (d^{-5})^{2}}{d^{-2}}
9. 1(xy)−3\frac{1}{(x y)^{-3}}

Studdy Solution

STEP 1

Assumptions1. All variables are non-zero. . The expressions need to be simplified and expressed with positive indices.

STEP 2

Let's start with the first expression. We'll use the power of a power rule, which states that (am)n=amn(a^m)^n = a^{mn}.
\left( w^{6}\right)^{} =^ \cdot (w^6)^

STEP 3

Calculate the power of3 and apply the power of a power rule to w6w^6.
(3w6)3=27⋅w6⋅3\left(3 w^{6}\right)^{3} =27 \cdot w^{6 \cdot3}

STEP 4

implify the expression.
(3w6)3=27w18\left(3 w^{6}\right)^{3} =27w^{18}

STEP 5

Now, let's move to the second expression. Again, we'll use the power of a power rule.
(22s5)3=(22)3⋅(s5)3\left(2^{2} s^{5}\right)^{3} = (2^2)^3 \cdot (s^5)^3

STEP 6

Calculate the power of2 and apply the power of a power rule to s5s^5.
(22s5)3=22⋅3⋅s5⋅3\left(2^{2} s^{5}\right)^{3} =2^{2 \cdot3} \cdot s^{5 \cdot3}

STEP 7

implify the expression.
(22s5)3=64s15\left(2^{2} s^{5}\right)^{3} =64s^{15}

STEP 8

For the third expression, we'll use the power of a fraction rule, which states that (ab)n=anbn(\frac{a}{b})^n = \frac{a^n}{b^n}.
(pq5)3=p3(q5)3\left(\frac{p}{q^{5}}\right)^{3} = \frac{p^3}{(q^5)^3}

STEP 9

Apply the power of a power rule to q5q^5.
(pq5)3=p3q5⋅3\left(\frac{p}{q^{5}}\right)^{3} = \frac{p^3}{q^{5 \cdot3}}

STEP 10

implify the expression.
(pq5)3=p3q15\left(\frac{p}{q^{5}}\right)^{3} = \frac{p^3}{q^{15}}

STEP 11

For the fourth expression, we'll again use the power of a fraction rule.
(−m3n5)5=(−m3)5(n5)5\left(\frac{-m^{3}}{n^{5}}\right)^{5} = \frac{(-m^3)^5}{(n^5)^5}

STEP 12

Apply the power of a power rule to -m^ and n5n^5.
(−mn5)5=−m⋅5n5⋅5\left(\frac{-m^{}}{n^{5}}\right)^{5} = \frac{-m^{ \cdot5}}{n^{5 \cdot5}}

STEP 13

implify the expression.
(−m3n5)5=−m15n25\left(\frac{-m^{3}}{n^{5}}\right)^{5} = -\frac{m^{15}}{n^{25}}

STEP 14

For the fifth expression, we'll use the rule that a−n=ana^{-n} = \frac{}{a^n}.
\left(s^{3}\right)^{-} \times s^{7} = \frac{}{(s^3)^} \times s^7

STEP 15

Apply the power of a power rule to s3s^3.
(s3)−5×s7=s3⋅5×s7\left(s^{3}\right)^{-5} \times s^{7} = \frac{}{s^{3 \cdot5}} \times s^7

STEP 16

implify the expression.
\left(s^{3}\right)^{-5} \times s^{} = \frac{s^}{s^{15}}

STEP 17

Use the rule that aman=am−n\frac{a^m}{a^n} = a^{m-n}.
(s3)−5×s7=s7−15\left(s^{3}\right)^{-5} \times s^{7} = s^{7-15}

STEP 18

implify the expression.
(s3)−5×s7=s−8\left(s^{3}\right)^{-5} \times s^{7} = s^{-8}

STEP 19

For the sixth expression, we'll use the rule that a−n=1ana^{-n} = \frac{1}{a^n} and aman=am−n\frac{a^m}{a^n} = a^{m-n}.
u^{-4} \div\left(u^{5}\right)^{-} = \frac{1}{u^4} \div \frac{1}{(u^5)^}

STEP 20

Apply the power of a power rule to u5u^5.
u−4÷(u5)−=u4÷u5⋅u^{-4} \div\left(u^{5}\right)^{-} = \frac{}{u^4} \div \frac{}{u^{5 \cdot}}

STEP 21

implify the expression.
u−4÷(u5)−=1u4÷1u10u^{-4} \div\left(u^{5}\right)^{-} = \frac{1}{u^4} \div \frac{1}{u^{10}}

STEP 22

Use the rule that abc=a⋅c\frac{a}{\frac{b}{c}} = a \cdot c.
u−4÷(u5)−=1u4⋅u10u^{-4} \div\left(u^{5}\right)^{-} = \frac{1}{u^4} \cdot u^{10}

STEP 23

Use the rule that aman=am−n\frac{a^m}{a^n} = a^{m-n}.
u−÷(u5)−=u10−u^{-} \div\left(u^{5}\right)^{-} = u^{10-}

STEP 24

implify the expression.
u−4÷(u)−=u6u^{-4} \div\left(u^{}\right)^{-} = u^{6}

STEP 25

For the seventh expression, we'll use the rule that a−n=1ana^{-n} = \frac{1}{a^n} and aman=am−n\frac{a^m}{a^n} = a^{m-n}.
\frac{d^{3} \div\left(d^{-5}\right)^{}}{d^{-}} = \frac{d^3 \div \frac{1}{(d^{-5})^}}{\frac{1}{d^}}

STEP 26

Apply the power of a power rule to d−5d^{-5}.
\frac{d^{3} \div\left(d^{-5}\right)^{}}{d^{-}} = \frac{d^3 \div \frac{1}{d^{-5 \cdot}}}{\frac{1}{d^}}

STEP 27

implify the expression.
\frac{d^{3} \div\left(d^{-5}\right)^{}}{d^{-}} = \frac{d^3 \div \frac{1}{d^{-10}}}{\frac{1}{d^}}

STEP 28

Use the rule that abc=a⋅c\frac{a}{\frac{b}{c}} = a \cdot c.
\frac{d^{3} \div\left(d^{-5}\right)^{}}{d^{-}} = d^3 \cdot d^{10} \cdot d^

STEP 29

Use the rule that am⋅an=am+na^m \cdot a^n = a^{m+n}.
d÷(d−5)2d−2=d+10+2\frac{d^{} \div\left(d^{-5}\right)^{2}}{d^{-2}} = d^{+10+2}

STEP 30

implify the expression.
d÷(d−5)2d−2=d15\frac{d^{} \div\left(d^{-5}\right)^{2}}{d^{-2}} = d^{15}

STEP 31

For the eighth expression, we'll use the rule that a−n=1ana^{-n} = \frac{1}{a^n}.
\frac{1}{(x y)^{-}} = \frac{1}{\frac{1}{(x y)^}}

STEP 32

Use the rule that abc=a⋅c\frac{a}{\frac{b}{c}} = a \cdot c.
\frac{1}{(x y)^{-}} =1 \cdot (x y)^

STEP 33

implify the expression.
\frac{1}{(x y)^{-}} = (x y)^The simplified expressions are11. 27w1827w^{18}
12. 64s1564s^{15}
13. \frac{p^}{q^{15}}
14. −m15n25-\frac{m^{15}}{n^{25}}
15. s−8s^{-8}
16. u6u^{6}
17. d15d^{15}
18. (x y)^

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