# $y=k^x$ when $k$ is negative

Hi I am learning about exponential functions and it points out that $$k$$ must be positive in $$y=k^x$$. I was wondering what would happen if $$k$$ is negative as I have drawn a rough copy of the graph and it seems quite interesting. Is it a valid graph and can anyone draw a stimulation of it please as I can't find an online graph sketcher that will draw it. Would $$y=(-1)^x$$ be almost like a oscillating wave?? Also can $$k=0$$ ? I'm just curious if this sort of graph exists. Thanks

• The outputs will only be real for certain values of x. For instance, it’s percectly valid to consider your function for integer values of x, but what happens when x is, say, 1/2? Feb 21, 2019 at 15:28

Lets consider $$k^x$$ when x is an integer and $$k<1$$.

Provided $$x$$ is an integer $$x \in \mathbb{I}$$ we have no issues.

Taking for example $$k = -2$$ we have:

$$\begin{array}{c|c|c} x&\text{}&k^x\\ \hline 3&(-2)^3&-8\\ 2&(-2)^2&4\\ 1&(-2)^1&-2\\ 0&(-2)^0&1\\ -1&\dfrac{1}{(-2)^1}&-\dfrac{1}{2}\\ -2&\dfrac{1}{(-2)^2}&\dfrac{1}{4}\\ -3&\dfrac{1}{(-2)^3}&-\dfrac{1}{8} \end{array}$$

Now consider $$x = \dfrac{1}{2}$$, and $$k^x = \sqrt{-2} = ?$$

We can solve this with complex numbers $$k^x = i \cdot \sqrt{2}$$

Where $$i = \sqrt{-1}$$

Similarly $$x = -\dfrac{1}{2}$$, and $$k^x = \dfrac{1}{\sqrt{-2}} = - \dfrac{i}{\sqrt{2}}$$

But our solutions are not real numbers any more $$k^x \notin \mathbb{R}$$

When $$k \lt 0$$ we do not have real solutions for all x

In the real plane, it doesn’t make sense for any $$x$$ that is not an odd integer.

However, if you were to draw it on an Agrand diagram, which is the complex plane, you’ll get a unit circle centered at the origin.

You can write $$-1=e^{i\pi}$$, so $$(-1)^x=e^{i\pi x}$$. This traces out the unit circle for $$x\in\mathbb{R}$$.

For negative $$k$$, you’ll get a spiral that spirals out anti-clockwise.

The problem is that $$k^x$$ is defined as $$\exp(x\log(k))$$ (at least if $$x$$ is not rational), which is problematic for negative $$k$$. Even if you restrict yourself to rational $$x$$, you need to explain what $$k^{1/2} = \sqrt{k}$$ is (and similarly for higher roots). It is only unproblematic for integer $$x$$.

• okay so is the graph not possible because you can't sqr root a negative number?
– yt.
Feb 21, 2019 at 15:31
• Do you really mean "The problem is that $k^x$ is defined as $\exp(x\log(k))$"? Perhaps you mean "is equal to"? Feb 21, 2019 at 15:32
• @yt. Yes, basically. And things get worse if you take $x$ irrational. Feb 21, 2019 at 15:34
• @user1729 How would you define $k^x$ for $x$ irrational? Feb 21, 2019 at 15:34
• I don't know. But I also do not know how to define $e^x$ for $x$ irrational. Feb 21, 2019 at 15:42