Inverting $f(x)=\frac{a^x-1}{a^x +1}$

Question

This is the problem:

$$f(x)= \frac{a^x-1}{a^x+1}, \quad a > 0, \quad a \ne 1.$$

What I can get, but I don't think it is right:

$$f^{-1}(x) = \frac{-x-1}{x \ln a-\ln a}.$$

So this is what I did:

$$x=\frac{a^y-1}{a^y+1}$$ $$xa^y+x=a^y-1$$ $$xa^y-a^y=-x-1$$ $$xy\ln(a)-y\ln(a)=-x-1$$ $$y(x\ln(a)-ln(a))$$ $$y=\frac{-x-1}{x\ln(a)-\ln(a)}$$

Solution:$$y=\frac{\ln\frac{-(x+1)}{x-1}}{\ln(a)}$$

Answer 1

You are fine down to $xa^y-a^y=-x-1$, but you take the log on the left and not on the right. When you do so, you assume the log of a sum is the sum of the logs, which is not correct. You seem to be claiming that $\ln (xa^y-a^y)=xy\ln a - y\ln a$. You should distribute out the $a^y$ to get$$(x-1)a^y=-x-1\\a^y=-\frac{x+1}{x-1}$$ now take logs and you are set.

Answer 2

Notice, let $f(x)=y\implies x=f^{-1}(y)$

$$y=\frac{a^x-1}{a^x+1}$$ $$ya^x+y=a^x-1$$ $$a^x(1-y)=1+y$$ $$a^x=\frac{1+y}{1-y}$$ $$\ln(a^x)=\ln\left(\frac{1+y}{1-y}\right)$$ $$x\ln a=\ln\left(\frac{1+y}{1-y}\right)$$ $$x=\frac{\ln\left(\frac{1+y}{1-y}\right)}{\ln a}$$$$\implies f^{-1}(y)=\frac{\ln\left(\frac{1+y}{1-y}\right)}{\ln a}$$ $$\color{red}{f^{-1}(x)=\frac{\ln\left(\frac{1+x}{1-x}\right)}{\ln a}=\log_a\left(\frac{1+x}{1-x}\right)}$$ $\forall \ \ a>0,\ a\ne 1$

Answer 3

Given $$\displaystyle \frac{f(x)}{1} = \frac{a^x-1}{a^x+1}\;,$$ Where $a>0\;,a\neq 1$

Now Using Componendo -Dividedno, We get

$$\displaystyle \frac{f(x)+1}{f(x)-1} = \frac{(a^x-1)+(a^x+1)}{(a^x-1)-(a^x+1)} = \frac{a^x}{-1}$$

So $$\displaystyle \frac{a^x}{1} = \frac{1+f(x)}{1-f(x)}$$

Now $$\displaystyle \ln_{a}(a)^x = \ln\left(\frac{1+f(x)}{1-f(x)}\right)\Rightarrow x = \ln\left(\frac{1+f(x)}{1-f(x)}\right)\;,$$

Now Using Definition of Inverse, Replace $x\rightarrow f^{-1}(x)$ and $f(x)\rightarrow x$

So we get $$\displaystyle f^{-1}(x) = \ln\left(\frac{1+x}{1-x}\right)\;,$$ Where $a>0$ and $a\neq 1$

Answer 4

You want to invert $y= \frac{a^x-1}{a^x+1} $.

Let $z = a^x$. Then you want to invert $y = \frac{z-1}{z+1} $. $zy+y = z-1$, so $z(1-y) =1+y$ or $z = \frac{1+y}{1-y} $.

Finally, $a^x= \frac{1+y}{1-y} $ or $x \ln a =\ln(1+y)-\ln(1-y) $.