# Laws of Exponents if base(s) negative

In the case where $$a,b>0$$ and $$x,y \in \mathbb R$$, we have Laws of Exponents such as these:

1. $$a^xa^y=a^{x+y}$$
2. $$(ab)^x=a^xb^x$$
3. $$a^{-x}=1/a^x$$
4. $$(a^x)^y=a^{xy}$$
5. $$a^x/a^y=a^{x-y}$$
6. $$(a/b)^x=a^x/b^x$$
7. $$(a/b)^{-x}=b^x/a^x$$

If instead $$a,b<0$$, then

• Which of the above Laws still hold?

• Which don't? (And how can they be modified to become true?)

This seems simple enough to check, so just go through each of the identities so that $$a, b < 0$$. Let $$p = -a$$ and $$q = -b$$, ($$p$$ and $$q$$ are clearly positive reals). Then, we can substitute these into the identities:

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

Substituting yields $$(pq)^x = (-p)^x(-q)^x$$. As was stated by J. W. Tanner in the comments, this is slightly meaningless because the left-hand side is a function that has only one output, which is real, and the right-hand side can potentially have multiple and imaginary outputs, specifically when $$x$$ is fractional with an even denominator or irrational. Therefore, for this one the law doesn't really hold, but could be modified to state that if $$x$$ is rational with an odd denominator in fully simplified form, then it would hold.

1. $$a^xa^y = a^{x+y}$$

Substituting yields $$(-p)^x(-p)^y = (-p)^{x + y}$$. Since $$(-p)^x$$ and $$(-p)^y$$ aren't necessarily uniquely defined, if we make the same assumptions as in the last case for $$x$$ and $$y$$ we can see that by rule 2 then we have: $$(-1)^x(p)^x(-1)^y(p)^y = (-1)^{x+y}(p)^{x+y}$$ Which becomes: $$(-1)^x(-1)^y(p^{x+y}) = (-1)^{x+y}(p)^{x+y}$$ $$(-1)^x(-1)^y = (-1)^{x+y}$$ At this point, it should be clear that for all of the identities, if we show that the identity holds for $$a,b = -1$$, it holds for all simply by applying rule 2, (assuming that the $$x,y$$ satisfy the previously mentioned restrictions. This is true because by applying rule 2 we can break any $$(-s)^x$$ into $$(-1)^x(s)^x$$, (again, meeting the restrictions). Therefore, we just need to check all the rules for $$-1$$.

Although it is not expressed here, quickly checking each of the equations for $$a,b = -1$$ reveals they are all true, again assuming that $$(-1)^x$$ is uniquely defined. So now to answer the second question: how can they be changed to work for negative values? Well then the only restriction is that $$(-1)^x$$ must be uniquely defined. I believe the link in the second comment by J. W. Tanner has more information on that.

Surprisingly, all of them hold. You just have to note when the powers are even because that leads to something like $$(-1)^2=1$$.

But I guess if you can show these exist of $$-1$$ and another random negative integer like $$-5$$ then proof's all good. But I'm leaving $$x, y \in \mathbb{N}$$ because the rules change magnificently when we walk out of our natural field. I'll show one example later on.

By the way, this is an example that goes to show simple ideas can extend to higher things if only they were defined well. Let's begin.

$$(-1)^3 \cdot (-5)^3=-1 \cdot -125 = 125 =(-1 \cdot -5)^3$$

$$(-1)^{2019}=(-1)^{2018} \cdot -1=(-1)^{2018+1}$$

That last one may seem to stem from some contest math but for a number $$k \in \mathbb{N}$$, $$(-1)^{2k}=[(-1)^2]^k$$. Boom! The fourth law just popped up. It just shows an even product of (-1)s can be arranged in groups of two of these, which become 1 i.e.

$$-1 \cdot -1 \cdot -1 \cdot -1=(-1 \cdot -1) \cdot (-1 \cdot -1)$$

This is a very helpful fact to note in future, especially if you ever do number theory. Let's move on.

$$(-5)^{-3}=(-1)^{-3} \cdot 5^{-3} = \dfrac{1}{5^3} \cdot \dfrac{(-1)^0}{(-1)^3}=\dfrac{1}{(-5)^3}$$

$$(-5)^2 \div (-5)=\dfrac{(-1)^2 \cdot 5^2}{-1 \cdot 5}=\dfrac{-1 \cdot 5^{2-1}}{1}=(-5)^1=(-5)^{2-1}$$

$$\dfrac{(-5)^x}{(-1)^x}=(-5)^x \cdot [(-1)^{-1}]^x=[-5 \cdot (-1)^{-1}]^x=(\dfrac{-5}{-1})^x$$

$$(\dfrac{-5}{-10})^{-x}=[\dfrac{-5}{-10}^x]^{-1} \qquad \text{Which is the reciprocal of the case that preceded it.}$$

So there you have it. Now for that special case.

In some textbooks Law. 2 is converted into something like:

$$\sqrt{a} \cdot \sqrt{b}=\sqrt{ab} \quad \text{Where in this case } x=0.5$$

Both a and b can't be negative because the equation can be shown to produce two separate numbers. Let's say $$-16,-36$$

$$\sqrt{a} \cdot \sqrt{b}=\sqrt{-16} \cdot \sqrt{-36}=\sqrt{-1} \cdot 4 \cdot \sqrt{-1} \cdot 6=-1 \cdot 24$$

$$\sqrt{ab}=\sqrt{-16 \cdot -36}=\sqrt{16 \cdot 36}=4 \cdot 6$$

One is positive and the other is negative. So that's the only exception. Peace.

• “Surprisingly, all of them hold”. The OP is asking which laws hold when $x,y \in \mathbb{R}$. Making the restrictiction $x, y \in \mathbb{N}$ is a non-answer, IMO. – Radial Arm Saw Jul 14 '20 at 0:52