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I am very confused on this question.

How does $(4^{-3} \cdot 2^{-3})^0 = (8^{-3})^0$? Should it not be $8^{-6}$ since $-3+-3 = -6$?

I just don't get where the other power of $-3$ goes off to?

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kingW3
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Mobius-User
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    For $a,b> 0$, $a^cb^c=(ab)^c$ – Hagen von Eitzen Jun 15 '17 at 11:54
  • Lets try $4^3\cdot 2^3$ you have $4\cdot 4\cdot 4\cdot 2\cdot 2\cdot 2$ this is equal to $(4\cdot 2)\cdot(4\cdot 2)\cdot(4\cdot 2)=8\cdot 8\cdot 8=8^3$. – kingW3 Jun 15 '17 at 12:01
  • That explains it quite well! thank you. However, i am now not sure what I would do in the case where there is not the same amount of powers. For instance, 4^3 * 3^7 What would i do then? – Mobius-User Jun 15 '17 at 12:09
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    @Novantix With $(4^3\cdot 3^7)^0$ you would skip steps 1 and 2 completely and note immediately that anything to the $0$th power is $1$. – Hagen von Eitzen Jun 15 '17 at 12:10

3 Answers3

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The rule you need to use is:

$$a^xb^x=(ab)^x$$

Consider $6^4\times 7^4$.

We could write this as \begin{align}6^4\times 7^4&=(6\times 6\times 6\times 6)\times (7\times 7\times 7\times 7)\\ &=(6\times7)\times(6\times7)\times(6\times7)\times (6\times7)\\ &=(6\times7)^4\\ &=42^4\end{align}

Can you now see why $$4^{-3}\times 2^{-3}=8^{-3}\neq8^{-6}$$

If the powers are different, then we can do the following:

\begin{align}6^2\times 8^5&=(6\times 6)\times(8\times 8\times 8\times8\times 8)\\ &=(6\times 8)\times (6\times 8)\times 8\times 8\times 8\\ &=(6\times 8)^2\times 8^3\\ &=48^2\times 8^3\end{align}

The general rule for this would be, for $x<y$ $$a^xb^y=(ab)^x\times b^{y-x}$$

However, this is not a usual method of simplifying, as it still leaves you with two numbers raised to different powers multiplied together

lioness99a
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  • ahh I see, thanks! What would you do in the case of 6^4 * 7^11? because then there would not be enough 6's to go with the 7. Would it then be in excess like: (6*7^7)^4 ? – Mobius-User Jun 15 '17 at 12:05
  • So basically, I could keep the terms separate like 6^4 * 7^4 However, if I want to then simplify it and turn it into one term, I have to see how many times that one term multiplies together, and that becomes the new power? so the new power would be 42, and I have to check and see how many times the 7 and 6 multiply together to create that power and depending on how many times, that would give me my power. This only works if they have the same amount of times that they multiply together to make that one term? – Mobius-User Jun 15 '17 at 12:11
  • We can check the value of $(6\times7^7)^4$ and see it is not equal. We could write $6^4\times 7^{11} = (6\times 7)^4\times 7^7$ as there would be $4$ lots of $(6\times 7)$ and then seven $7$s left over. – lioness99a Jun 15 '17 at 12:12
  • That is a perfect explanation, I understand now how it all works! :) – Mobius-User Jun 15 '17 at 12:14
  • @Novantix Thank you, if you feel it has answered your question, you can click the tick next to it to mark it as the best answer (and we will both get some reputation) – lioness99a Jun 15 '17 at 12:15
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Remember that for every $x \neq 0$ we have that $x^0 = 1$.

Hans Hüttel
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You seem to think that $a^xb^x = (ab)^{x+x}=(ab)^{2x} $ or perhaps even $a^xb^y = (ab)^{x+y}$, but the rule you need in that step is: $$a^xb^x = (ab)^x$$ So: $4^{-3}2^{-3} = \left( 4 \cdot 2 \right)^{-3} = 8^{-3}$.

StackTD
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