# Shifted output of input vs output of shifted input

As I understand it, shifted output means first getting the output and then shifting it. We feed in $$x[n]$$, and get our output as $$x[2n]$$. Then we shift it by $$k$$ to get $$x[2n-k]$$. This is the shifted output.

Now, for output of the shifted input, we shift the input to get $$x[n-k]$$, then feed it as input to get output as $$x[2(n-k)]=x[2n-2k]$$

However, the text in the image has these answers reversed. Is there something I'm not getting?

This is all about consequently substituting variables. Shifting by $$t_0$$ means replacing the variable $$t$$ by $$t-t_0$$. If $$y(t)=x(2t)$$, then $$y(t-t_0)=x(2(t-t_0))=x(2t-2t_0)$$.
If you have a shifted input $$x_2(t)=x(t-t_0)$$, then the output is $$y(t)=x_2(2t)$$, i.e., the variable $$t$$ is multiplied by $$2$$. Replacing $$t$$ by $$2t$$ in $$x(t-t_0)$$ results in $$x(2t-t_0)$$.
EDIT: Concerning shifting the output, imagine a triangular input $$x(t)$$ with its apex at $$t=0$$. The output is a compressed version of the input, i.e., it is also triangular with its apex at $$t=0$$. If the output is shifted by $$t_0$$ the apex is shifted to $$t=t_0$$. The function $$x(2t-t_0)$$ has its apex at $$t$$ satisfying $$2t-t_0=0$$, i.e., at $$t=t_0/2$$, so the correct formula for the shifted output is $$x(2t-2t_0)$$.