Question 1006363: pleace. help me the solution
for this question
Inverse f(x)=2+ln(x-1)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! let y = f(x).
your equation becomes y = 2 + ln(x-1).
this can be written as:
y = ln(x-1) + 2
to find the inverse, do the following:
replace y with x and x with y to get:
x = ln(y-1) + 2
believe it or not, this is your inverse eqaution.
all it requires is to replace x with y and y with x.
the rest is solving for y which, technically, is not necessary, but is useful if you're graphing the equation and using software that is not capable of graphinjg it as is.
so you start with:
x = ln(y-1) + 2
subtract 2 from both sides of the equation to get:
x - 2 = ln(y-1)
this can be written as:
ln(y-1) = x - 2
by the basic definition of logs, this is true if and only if:
e^(x-2) = y - 1
add 1 to both sides of this equation to get:
e^(x-2) + 1 = y
this can be written as:
y = e^(x-2) + 1
that's your inverse equation aftger solving it for y.
since y = f(x), this can also be written as:
f(x) = e^(x-2) + 1
the graph of the original equation of y = ln(x-1) + 2 and the graph of the inverse equation of y = e^(x-2) + 1 are shown below.
to prove that they are inverse equations of each other, they have been shown along with the graph of y = x and the distances from select points on each graph have been compared with reach other through intersections with the line y = -x.
you get (x,y) on one equation is equal to (y,x) on the other equation.
this confirms they are inverse equations of each other.
the original equation is in black.
the inverse equation is in red.
the lines y = x and y -x ... are in dashed green.
to prove that the inverse equation of y = ln(x-1) + 2 is found by simply interchanging y and x, then the following graph shows the equation of y = ln(x-1) + 2 and the equation of x = ln(y-1) + 2.
this can be done because the desmos calculator is capable of graphing both ways.
as you can see, the graph of x = ln(y-1) + 2 is identical to the grpah of y = e^(x-2) + 1
they are the exact same equations, only one has been solved for x and the other has been solved for y.
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