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# Explain why (–4x)^0 = 1, but –4x^0 = –4.

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Explain why (–4x)^0 = 1, but –4x^0 = –4.

Jul 7, 2015

#2
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Explain why (–4x)^0 = 1, but –4x^0 = –4.

I. (–4x)^0 = 1

$$1=\dfrac{(-4x)^1}{(-4x)^1}=(-4x)^1\cdot (-4x)^{-1}=(-4x)^{1-1}=(-4x)^0$$

II. –4x^0 = –4

$$-4 = -4\cdot 1 = -4 \cdot \dfrac { x^1 } { x^1 } = -4 \cdot x^1 \cdot x^{-1} = -4 \cdot x^{1-1} = -4 \cdot x^0$$

Jul 7, 2015

#1
+423
+10

This is due to a readability issue. You see, normally you follow an order of significance: first you do powers, then you do multiplications, then additions (and the inverse of each happens at the same time, and goes from left to right).

In the first case, you got (-4 * x)^0

Well anything to the power of 0 = 1 so clearly this equation is: (-4 * x)^0 = 1

In the second case, you got -4 * x ^0

The way this must be read is doing the powers first, so we get -4 * 1

This then multiplies and gives us: -4 * x ^0 = -4

Remember: it never hurts to use more parenthesies if you really want to make it clear what order you meant. It's important to make sure the equation is easy to read.

Jul 7, 2015
#2
+20831
+10

Explain why (–4x)^0 = 1, but –4x^0 = –4.

I. (–4x)^0 = 1

$$1=\dfrac{(-4x)^1}{(-4x)^1}=(-4x)^1\cdot (-4x)^{-1}=(-4x)^{1-1}=(-4x)^0$$

II. –4x^0 = –4

$$-4 = -4\cdot 1 = -4 \cdot \dfrac { x^1 } { x^1 } = -4 \cdot x^1 \cdot x^{-1} = -4 \cdot x^{1-1} = -4 \cdot x^0$$

heureka Jul 7, 2015