Why is $(\frac{x}{y})^x=n^x$ linear when graphed. I am currently learning the end of Algebra $1$. I was experimenting in demos and found that $(\frac{x}{y})^x=n^x$ is a linear equation. Furthermore, $n$ functions as $m$ would in $y=mx+b$ with. Why is this?
 A: A much simpler way is to just consider the equation
$$x^2=2^2$$
To solve for $x$ we take the square root of both sides. One way of thinking of this equation is
$$\left(\frac{x}{y}\right)^x=n^x$$
and in a similar fashion, we take the $x$ root of both sides
$$\sqrt[x]{\left(\frac{x}{y}\right)^x}=\sqrt[x]{n^x}$$
which then cancels the $x$. giving
$$\frac{x}{y}=n\implies y=\frac{x}{n}$$
Which is the equation of a straight line $y=mx+b$ with $b=0$ and $m=1/n$
A: Maybe using the log makes it more obvious, (for $y\ne 0$, $\left(\dfrac{x}{y}\right)^x > 0$, $n^x > 0$ )
$$\log\left(\left(\dfrac{x}{y}\right)^x\right) = \log\left(n^x\right) $$
since $\log(a^b) = b\cdot \log(a)$
$$x\cdot \log\left(\dfrac{x}{y}\right) = x\cdot \log(n)$$
simplify the $x$, and since $e^{\log(x)}=x$
$$\dfrac{x}{y} = n$$
we get the usual linear $y = ax + b$ form
$$y = \dfrac{1}{n}\cdot x + 0$$
which is linear.
A: $1^x$ is always $1$, so you're just graphing $(\frac xy)^x=1$.
A number to any power is $1$ if and only if the number itself is $1$, so you're really just graphing $\frac xy=1$.
And that's just $x=y$.
A: Let us assume that $x \not =0$. Then there is a nonzero value of $y$ that solves this equation. Multiplying both sides by $y^x$ then gives $$x^x = 1^xy^x =y^x,$$ and so this equation is solved if the equation $x=y$ holds.
A: It's actually not a function. It's a straight line $y=x$, except $x=2t$, $y=\pm 2t$ are solutions for all $t \in \mathbb{Z}$. Think of it like the line $y=-x$ is also a solution but only when $x$ is an even integer and $y$ is negative $x$.
