Hello, geton!
Sean is right . . . There must be dozens of approaches.
Prove that if ab)^{xy}" alt="a^x \:= \:b^y \:= \ab)^{xy}" />, then: .
We have: . ab)^{xy}" alt="a^x \:=\ab)^{xy}" />
Take logs: .
* Divide by .[1]
We have: . ab)^{xy}" alt="b^y \:=\ab)^{xy}" />
Take logs: .
* Divide by .[2]
Add [1] and [2]: .
. . Therefore: .
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*
If , we have: . ab)^0\quad\Rightarrow\quad b^y \:=\:1" alt="a^0 \:=\:b^y \:=\ab)^0\quad\Rightarrow\quad b^y \:=\:1" />
There are two possibilities: .
In case (1),
In case (2), is strictly determined which contradicts the original equation
. . which is true for all values of
Hence: .
Similarly, we can show that: .