$2^{n}$

$\begin{aligned} 2^n - 2^{n-1} & = 2^x \\ 2^{n-1}(2-1) & = 2^x \\ 2^\red{n-1} & = 2^\red x \\ \implies x & = \boxed{n-1} \end{aligned}$

$\begin{aligned} 2^n - 2^{n-1} &= 2^x \\ 2^{n-1}(2-1) &= 2^x \\ 2^{n-1} &= 2^x \\ \log_2 (2^{n-1}) &= \log_2 (2^x) \\ n-1 &= x \end{aligned}$

We first write out the original equation.

$2^n-2^{n-1}=2^?$

Then, we write $2^n$ as $2*2^{n-1}$

$(2-1)*2^{n-1}=2^?$

$2^{n}=2^{?}$

FInally, we see that the ? is equal to n-1, so $\boxed{n-1}$ is the answer.