Let f(x)=20/(1+3e^-x). Find a formula for f^-1(x).
So I tried to do it this way:
y=f(x)=20/(1+3e^-x).
x=20/(1+3e^-y)
20/x=1+3e^-y
log ((20/x)-1)/log(3e)
Does this look right, or am I going about this in the wrong way?
Let f(x)=20/(1+3e^-x). Find a formula for f^-1(x).
So I tried to do it this way:
y=f(x)=20/(1+3e^-x).
x=20/(1+3e^-y)
20/x=1+3e^-y
log ((20/x)-1)/log(3e)
Does this look right, or am I going about this in the wrong way?