Special cases of the Laws of Exponents?

I have a few questions regarding powers.

We have:

\begin{align} (x\cdot y)^n &= x^n\cdot y^n \tag 1\\ (x^m)^n &= x^{m\cdot n} \tag 2\\ x^m \cdot x^n &= x^{m+n} \tag 3 \end{align}

Question 1:

I guess a special case of $$(1)$$ is $$y=x$$, so: \begin{align} (x\cdot x)^n \underbrace{=}_\text{(1)} x^n\cdot x^n \underbrace{=}_\text{(3)} x^{n+n} =x^{2n} \tag 4 \end{align} Is this correct?

Question 2:

If a special case of $$(3)$$ is $$m=n$$ we have: $$x^n\cdot x^n \underbrace{=}_\text{(3)} x^{n+n}=x^{2n} \tag 5$$

Is this correct?

But if $$m=n$$ in $$(2)$$ we have: $$(x^n)^n \underbrace{=}_\text{(2)} x^{n\cdot n} = x^{n^2} \tag 6$$

And this doesn't make sense... I guess it contradicts $$(4)$$ and $$(5)$$?

• Why does the last one not make sense? – blub Apr 3 at 9:34
• Everything seems in order for me ... – Matti P. Apr 3 at 9:35
• Hi! I updated the question. I think $(6)$ contradicts $(4)$ and $(5)$? – Joyat Apr 3 at 9:37
• I think you accidently assume $(x^n)^n = x^n \cdot x^n$ which is of course not corret. – The Pheromone Kid Apr 3 at 9:38
• You are probably thinking that $(x^{n})^{n}=x^{n}x^{n}$. That is not true. There is no contradiction in 4) ,5) and 6). – Kabo Murphy Apr 3 at 9:39

I think your confusion may be the fact that you think you calculate the same expression three times, but in the first two cases you compute $$\color{blue}{x^n \cdot x^n}$$ in two ways (or using two properties) while in the last case, you compute $$\color{purple}{(x^n)^n}$$ and that's not the same: $$\underbrace{\color{blue}{x^n \cdot x^n}}_{2\text{ factors}} = (x^n)^2 \color{red}{\;\mathbf{\ne}\;} \color{purple}{(x^n)^n} = \underbrace{x^n \cdot x^n\cdot \ldots\cdot x^n}_{n\text{ factors}}$$

From the looks of it, you expected $$x^n\cdot x^n$$ and $$(x^n)^n$$ to be the same, somehow?

Well, they are not. If we expand the exponents into repeated products of $$x$$ (this is assuming $$n$$ is a natrural number), we see that $$x^n\cdot x^n = \underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{n\text{ times}}$$ for a total of $$2n$$ $$x$$'s, while $$(x^n)^n = (\underbrace{x\cdot x\cdots x}_{n\text{ times}})^n\\ = \underbrace{\underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdot \underbrace{x\cdot x\cdots x}_{n\text{ times}}\cdots\underbrace{x\cdot x\cdots x}_{n\text{ times}}}_{n\text{ times}}$$ for a total of $$n\cdot n$$ $$x$$'s.

Of course, if $$n = 2$$, or $$n = 0$$, or $$x = 1$$ or $$x = 0$$, then they happen to be the same. But in general they are quite different.

Question 1

This is indeed correct. In fact, another way to see it is to work from your equation (2): $$(x \cdot x)^n = (x^2)^n \underbrace{=}_{(2)} x^{2n}$$

Question 2

This is also correct.

Question 3

Why does this not make sense? Consider a simple example with $$x=2, m=2, n=3$$. Equation (2) provides $$(x^m)^n = x^{mn} = 2^{2 \cdot 3} = 2^6 = 64$$ If we work it out without using the identity, we obtain $$(x^m)^n = (2^2)^3 = 4^3 = 64$$ which is the same. If you think the identity does not make sense, then what would your expected result be? Try an example and see if it is correct.

I hope this helps!