Question #9e52a

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Question #9e52a

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Answer 1 · 2016-09-30T18:04:38

The product rule and the power rules in question are

$a^{m} \times a^{n} = a^{m + n}$ $a^mxxa^n = a^(m+n)$
$a^{m} \div a^{n} = a^{m - n}$ $a^m-:a^n = a^(m-n)$
${(a^{m})}^{n} = a^{m \times n}$ $(a^m)^n = a^(mxxn)$

(Note that the second rule comes directly from the first, because $\frac{1}{a^{x}} = a^{- x}$ $1/a^x = a^(-x)$ , so $\frac{a^{m}}{a^{n}} = a^{m} \times a^{- n} = a^{m + (- n)} = a^{m - n}$ $a^m/a^n = a^mxxa^(-n) = a^(m+(-n)) = a^(m-n)$ )

Rather than simply post answers to all of the questions on the worksheet, here are a couple of examples of each. These should show how to do the rest, as the problems are all effectively the same, save for the changed values.

First set:

1) $5^{- 8} \times 5^{- 5} = 5^{- 8 + (- 5)} = 5^{- 13}$ $5^(-8)xx5^(-5) = 5^(-8+(-5)) = 5^(-13)$

2) $18^{- 4} \times 18^{3} = 18^{- 4 + 3} = 18^{- 1}$ $18^(-4)xx18^(3) = 18^(-4+3)=18^(-1)$

Second set:

1) $4^{2} \div 4^{10} = 4^{2 - 10} = 4^{- 8}$ $4^2-:4^10 = 4^(2-10) = 4^(-8)$

2) $19^{7} \div 19^{- 8} = 19^{7 - (- 8)} = 19^{15}$ $19^7 -: 19^(-8) = 19^(7-(-8)) = 19^15$

Third set:

1) ${(15^{9})}^{- 7} = 15^{9 \times - 7} = 15^{- 63}$ $(15^9)^(-7) = 15^(9xx-7)=15^(-63)$

2) ${(7^{3})}^{6} = 7^{3 \times 6} = 7^{18}$ $(7^3)^6 = 7^(3xx6)=7^18$

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Question #9e52a

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