Question 1163140
this was a beauty, but i think i have it, even though it's not what i would have thought.


normally, when you solve for the inverse function, you do the following:


replace y with x and x with y and then solve for y.


for example, consider y = ln(x)
replace x with y and y with x to get x = ln(y)
this is true if and only if y = e^x.
the inverse function of y = ln(x) is y = e^x. *****
that can be seen in the following graph.


<img src = "http://theo.x10hosting.com/2020/080501.jpg" >


we'll do the same procedure with your problem.
your equation is y = ln(sqrt(e^(x-2))
replace x with y and y with x to get x = ln(sqrt(e^(y-2))
this is true if and only if e^(y-2) = e^x
this is true if and only if y-2 = x
solve for y to get y = x + 2 
your inverse equation is y = x + 2 *****
that can be seen in the following graph.


<img src = "http://theo.x10hosting.com/2020/080502.jpg" >


you will notice that both the graph of the original equation and the graph of the inverse equation are straight lines.


i figured if the original equation was a straight line and was shown in the form of a natural log function, then the inverse function, being a straight line, could also be shown in the form of a natural log funcion.


i ran out of time to explain this fully, but the results of my investigation was that the inverse function could also be shown in the form of a natural log function.


i got the following:


the original function is y = ln(sqrt(e^(x-2))
the inverse  function is y = ln(e^(2x+2))
that can be shown in the following graph.


<img src = "http://theo.x10hosting.com/2020/080503.jpg" >



take it for what it's worth.
your inverse function is either:
y = 2x + 2, or:
y = ln(e^(2x+2))


both of these can be seen on the graphs.
the graphs show them to be reflections about the line y = x and (x,y) = (y,x).
both of these are indications of inverse functions.


i'm actually surprised i was able to see the relationship, but it's there, even it i can't explain if very well.