What does this rule mean?

I read this rule in a book .it says, If $$a>0$$, $$a$$ is not equal to $$1$$ and $$a^x=a^y$$,then $$x=y$$. But I don't understand why the value of $$a$$ has to be greater than $$0$$. What if the value of $$a$$ was less than $$0$$? Wouldn't it be the same ?For example, if $$a=-2$$, for which other value of $$x$$,could I get $$4$$ other than $$2$$?

• You don't define a real exponential function $a^x$ for $a<0$. Just imagine the domain of that!!
– Pspl
Nov 6 '19 at 15:12
• Could you please explain what you mean to say? I am new to these rules.Thanks Nov 6 '19 at 15:19
• Ok! Using your own terms: what is the domain of $a^x$ where $a=-2$? Can you tell?
– Pspl
Nov 6 '19 at 15:20
• A silly example as well, $(-1)^1 = (-1)^3$ despite $1\neq 3$. Of course, you could change your statement to say $|a|$ is not equal to $1$ to cover this case. That said, even if you can't find another $y$ such that $(-2)^2 = (-2)^y$ that doesn't imply that for all $x,y$ if $(-2)^x=(-2)^y$ that it would imply $x=y$. Consider what happens when you start using fractions in the exponent instead. Nov 6 '19 at 15:28
• Could please give an example of what you mentioned in the last sentences? Nov 6 '19 at 16:52

When $$a<0$$ exponentiation is not well defined for $$x \in \mathbb R$$.
Note that for $$a>0$$, $$a\neq 1$$ the equality $$x=y$$ from $$a^x=a^y$$ holds since $$f(x)=a^x$$ is injective.
• Provided $a\neq1$. Nov 6 '19 at 15:20