Special cases of the Laws of Exponents?

Question

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 $(5)$?

Thanks in advance!

Answer 1

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: $$\underbrace{\color{blue}{x^n \cdot x^n}}_{2\text{ factors}} = (x^n)^2 \color{red}{\;\mathbf{\ne}\;} \color{purple}{(x^n)^n} = \underbrace{x^n \cdot x^n\cdot \ldots\cdot x^n}_{n\text{ factors}} $$

Answer 2

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.

Answer 3

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!