Negative exponents problem.

Question

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?

Answer 1

The rule you need to use is:

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

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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 $$4^{-3}\times 2^{-3}=8^{-3}\neq8^{-6}$$

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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

Answer 2

Remember that for every $x \neq 0$ we have that $x^0 = 1$.

Answer 3

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: $4^{-3}2^{-3} = \left( 4 \cdot 2 \right)^{-3} = 8^{-3}$.