The Inverse of a Exponential/Rational Equation

**chrozer** · November 30th 2008, 01:31 PM

Find a pattern for $f^{-1}(x)$ if
$f(x) = \frac {a^x +1}{a^x -1}$
where $a>0, a \neq 1$ .

I have no idea what the question is asking by "a pattern". Does it mean the limitations of the inverse?

I also tried to solve it by switching the "x" values with "y" values and working from there but I got stuck at $x(a^y - 1) = a^y +1$ .

**skeeter** · November 30th 2008, 01:40 PM

$y = \frac{a^x + 1}{a^x - 1}$

$x = \frac{a^y + 1}{a^y - 1}$

$xa^y - x = a^y + 1$

$xa^y - a^y = x + 1$

$a^y(x - 1) = x + 1$

$a^y = \frac{x+1}{x-1}$

$y = \log_a\left(\frac{x+1}{x-1}\right)$

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