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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 $2$?

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  • $\begingroup$ You don't define a real exponential function $a^x$ for $a<0$. Just imagine the domain of that!! $\endgroup$
    – Pspl
    Nov 6 '19 at 15:12
  • $\begingroup$ Could you please explain what you mean to say? I am new to these rules.Thanks $\endgroup$
    – user459284
    Nov 6 '19 at 15:19
  • $\begingroup$ Ok! Using your own terms: what is the domain of $a^x$ where $a=-2$? Can you tell? $\endgroup$
    – Pspl
    Nov 6 '19 at 15:20
  • $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Nov 6 '19 at 15:28
  • $\begingroup$ Could please give an example of what you mentioned in the last sentences? $\endgroup$
    – user459284
    Nov 6 '19 at 16:52
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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.

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  • $\begingroup$ Provided $a\neq1$. $\endgroup$ Nov 6 '19 at 15:20
  • $\begingroup$ @MichaelHoppe Yes of course! I add that. Thanks $\endgroup$
    – user
    Nov 6 '19 at 15:21

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