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$\dfrac{(8^3)(-16)^5}{4(-2)^8}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{4\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

$\dfrac{8\cdot8\cdot8\cdot-16\cdot-16\cdot-16\cdot-16\cdot-16}{2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2}$

is it equal to... ? $$ \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} $$

I am bit confused, How can I handle this problem ?

azimut
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Franck
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2 Answers2

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Hint: Just take everything to powers of 2 using the laws of exponents

$$\frac{8^3(-16)^5}{4(-2)^8}=\frac{-(2^3)^3(2^4)^5}{2^2\cdot2^8}=?$$

Ross Millikan
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  • I don't like this. It's really hard to get students to cancel before multiplying. Often they see $\dfrac{20\cdot19\cdot18\cdot17\cdot16\cdot15}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$ and they start by multiplying out the numerator and the denominator! – Michael Hardy Sep 12 '11 at 18:38
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    @Michael Hardy: to me it depends upon the problem. For this one, especially without a calculator, all those powers of 2 scream at me and I would like an answer of -(2^n) as I know those (not quite this high). I can do $5\cdot 4+3 \cdot 3-2-8$ in my head. – Ross Millikan Sep 12 '11 at 18:43
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When multiplying fractions, cancel BEFORE multiplying.

$$ \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{8}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} \cdot \dfrac{-16}{2} $$

Wherever you see $\dfrac{8}{2}$, put $4$.

Wherever you see $\dfrac{-16}{2}$, put $-8$.

Then you have $$ 4\cdot4\cdot4\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8)\cdot(-8). $$

Michael Hardy
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