Question 1125603
{{{matrix(1,3,
"f(x)", ""="",log(2,(10x-20)))}}}<pre>
<font size = 4><b>
That has the red graph, with the blue vertical asymptote:

{{{drawing(400,400,-2,5,-2,5,blue(line(2,-6,2,6)),
graph(400,400,-2,5,-2,5,ln(10x-20)/ln(2)))}}}


Replace f(x) by y

{{{matrix(1,3,
"y", ""="",log(2,(10x-20)))}}}

Interchange x and y

{{{matrix(1,3,
"x", ""="",log(2,(10y-20)))}}}

Solve for y.

Raise 2 to the both sides power:

{{{matrix(1,3,
2^x, ""="",2^(log(2,(10y-20))))}}}

Use a rule of logs {{{a^( log(a,(b))) =b}}} on the right side:

{{{matrix(1,3,
2^x, ""="",10y-20)}}}

{{{matrix(1,3,
-10y,""="",-2^x-20)}}}

Divide both sides by -10

{{{matrix(1,3,
(-10y)/(-10),""="",(-2^x)/(-10)-(20)/(-10))}}}

{{{matrix(1,3,
y,""="",2^x/10+2)}}}

Replace y by f<sup>-1</sup>(x)

f<sup>-1</sup> ={{{2^x/10^""+2}}}   <--- the answer!

This has the green graph with the blue horizontal asymptote:

{{{drawing(400,400,-2,5,-2,5,blue(line(-6,2,6,2)),
graph(400,400,-2,5,-2,5,34,2^x/10+2))}}}

Put the original red function together with its green inverse on the 
same set of axes:

{{{drawing(400,400,-2,5,-2,5,blue(line(-6,2,6,2)),blue(line(2,-6,2,6)),
graph(400,400,-2,5,-2,5,ln(10x-20)/ln(2)),
graph(400,400,-2,5,-2,5,34,2^x/10+2))}}}

Now if we draw the identity line (dotted) whose equation is y = x,

{{{drawing(400,400,-2,5,-2,5,blue(line(-6,2,6,2)),blue(line(2,-6,2,6)),
graph(400,400,-2,5,-2,5,ln(10x-20)/ln(2),12,13,x*sqrt(sin(9x))/sqrt(sin(9x))),
graph(400,400,-2,5,-2,5,34,2^x/10+2))}}}
 
we see that the green inverse function is the red original function
reflected across the dotted identity line which has equation y = x.

Edwin</pre></font></b>