Question 936359
y=2^(x+1)  is very different from y=2^x+1.


You want this equation:  y=2^(x+1).  Rendering makes it appear, {{{y=2^(x+1)}}}.


The quick way to get inverse,
Switch x and y roles.
Keeping in base 2...
{{{x=2^(y+1)}}}
{{{log((x))=log((2^(y+1)))}}}
{{{log((x))=(y+1)*log((2))}}}
{{{log((x))=(y+1)*1}}}
{{{y+1=log((x))}}}
{{{y=log((x))-1}}}, the equation for the inverse of your given exponential function.  BASE IS 2.


You can find a table of values for your given exponential function, since this is a very elementary level task.  You will find a horizontal limit of y=0.



The given exponential function graph, BASE IS 2.
{{{graph(250,250,-5,9,-2,14,2^(x+1))}}}


Domain and range switch for the inverse.
{{{graph(250,250,-5,9,-12,2,log(2,x-1))}}}
The system unfortunately is not fully displaying the inverse graph.  The asymptote should be understood and appear as y=0, a vertical asymptote.