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I need to prove it for $7$ points.
If $a^x = a^z$ then $x = z$. What's the name of this rule?
I need a proof.

Harnoor Lal
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Naser Kot
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  • Rule* nvm just correcting – Naser Kot Mar 06 '17 at 18:24
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    What type of object is $A$? Without more context it is false. Think about powers of one or powers of the identity matrix. – JMoravitz Mar 06 '17 at 18:27
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    Yuor claim fails if and only if either $0$ or a root of $1$ are eigenvalues of $A$. –  Mar 06 '17 at 18:32
  • Mmm I'm just a 14 years old kid . – Naser Kot Mar 06 '17 at 18:42
  • That's not an excuse. If you are old enough to be asked this question, you should be old enough to answer this question. – JMoravitz Mar 06 '17 at 18:43
  • it's just a number and the rule also states that it shouldn't be negative nor zero , I need the name or the proof please please – Naser Kot Mar 06 '17 at 18:44
  • You still haven't answered the question in my very first comment... what type of object is $A$? Is it a real number? Is it a matrix? Is it a function? Is it a pet cat? Are there any restrictions on it it? Again, it is false until you have specified further. $1^5=1^9$ but $5\neq 9$ – JMoravitz Mar 06 '17 at 18:45
  • Sorry , real number – Naser Kot Mar 06 '17 at 18:45
  • More like (1/2)^x = 2^8 solve for x .. that simple – Naser Kot Mar 06 '17 at 18:48
  • I need the name or the proof it's extra points just for the research even if we don't know how we've came to that result – Naser Kot Mar 06 '17 at 18:49
  • Until you've corrected the statement, I can't give a name to it except "false." There is no reason to give a name to a false statement. Once you've corrected the statement, the phrase "monotonicity of the exponential function" should help by noticing that for $A$ a real number strictly greater than $1$ you have $A^x=A^y$ if and only if $1=A^{y-x}$ – JMoravitz Mar 06 '17 at 18:52

2 Answers2

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This isn't a rule, at least as you've presented it. For example, $1^4 = 1^2$, but $4 \neq 2$. Similarly, this holds for $0^4 = 0^2$ and $(-1)^4 = (-1)^2$.

However, if $A$ is a positive number with $A \neq 0$ and $A \neq 1$, we have $$ A^x = A^z \implies \\ \log(A^x) = \log(A^z) \implies\\ x \log(A) = z \log(A) \implies\\ x = z $$ I don't know of any "name" for this rule. Perhaps we can call it the "well-definedness of the log base $A$", or "the injectivity of $f(x) = A^x$".

Ben Grossmann
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Before this question can be answered, you need to clarify where A is a real number, a 1-D array, or a 2-D array (matrix), etc. This is important because the power of a matrix does not act the same on a 2-D matrix as on a 1-D integer. Linear algebric operators have very specific meaning, therefore you need to be careful when defining your problem. For example, if A were the matrix [[1,2],[1,5]], then A^0=I, A to the zeroth power equals the identity matrix. This is different from a 1-D, real number. Let's assume A is matrix. If A is the identity, then, this statement does not hold because I^3=I^9, but 3!=9, and you have proven it not to be true by contradiction. Similarly, if A were the number 1, this would not be true either by contradiction. Hope this helps.

Andyman
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