Can $x^y=1$ be regarded as a pair of straight lines?

Question

For $x> 0$ can $x^y=1$ be regarded as a pair of straight lines $x =1$ and $y =0$. U can get this by taking log on both sides.

Answer 1

Sort of. If you mean "Does the solution set of $x^y = 1$ consist of two lines?", then the answer is "yes". As for regarding an equation as the same thing as its solution set...well, no.

Taking log of both sides is a little risky in general, but writing $$ x^y = e ^ {y \log x} $$ which is just the definition (sort of ... see below), then makes it possible to write $$ e^{y \log x } = e^0 $$ and since exponentiation is injective, you get $$ y \log x = 0. $$ whose solutions are just those of $y = 0$ and $\log x = 0$, and the latter has the same solution set as $x = 1$.

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The only tricky thing here is that $x^y$ is generally defined by $e ^ {y \log x}$ for real values of $y$ and for $x > 0$, but for integer values of $y$, some negative values of $x$ are allowable as well, so that for instance $$ (-1)^2 $$ is well-defined and is $1$, even though $e ^ {2 \log (-1)}$ is not defined, because $\log u$ makes sense only for positive $u$.

So to talk about the equation you wrote, you really need to be very clear about the domain you're considering, and whether any negative values of $x$ are being considered or not. (Even if they are, it turns out that $x^y$ will never be $0$...but for different reasons than the one I gave above.) Fortunately, in the question itself (although not in the title), you included the restriction that $x > 0$, so my solution works in that case.