I'll just start out by showing the equation.

$\displaystyle y = (10^x + 10^-^x) / (10^x - 10^-^x)$

I had a similar problem. I'll show how I worked that one out so that you can see my train of thought on this one.

$\displaystyle y = (10^x + 10^-^x)/2$

$\displaystyle 2y = 10^x + 10^-^x$

Multiply both sides by $\displaystyle 10^-^x$

$\displaystyle 2y10^x = 10^2^x +1$

Now it's a quadratic. Solving for $\displaystyle 10^x, a=1, b=-2y, and c=1$

$\displaystyle 10^2^x - 2y^x +1 = 0$

I'll save both of us some time by not showing the quadratic equation steps.

So, I now have:

$\displaystyle 10^x = y +- sqrt(y^2-1)$

Now, I take the log of both sides:

$\displaystyle log10^x = log(y +- sqrt(y^2-1))$

So,$\displaystyle x = log(y +- sqrt(y^2-1))$

Now, back to the problem at hand.

I multiplied both sides of the original equation by $\displaystyle 10^x - 10^-^x$ to get:

$\displaystyle y10^x - y10^-^x = 10^x - 10^-^x$

I then multiplied both sides by $\displaystyle 10^-^x$ to get:

$\displaystyle y10^2^x - y = 10^2^x - 1$

I then bring everything to one side:

$\displaystyle y10^2^x - 10^2^x - y + 1 = 0$

But, I can't see how this is quadratic. Either I screwed up before this point (the most likely scenario), or I'm just not seeing the quadratic.

Please help. Thanks.