# Solving (or estimating) $x$ in $\tau=\log_x\left(\frac{x+1}{2}\right)$

How would one find a real value for $x$ that satisfies

$$\tau=\log_x\left(\frac{x+1}{2}\right),$$

given $0 < \tau < 1$ and $\tau \neq \frac{1}{2}$

(PS I'm not that good with math, so if this is impossible, please explain it to me like I'm 5). I don't need an exact solution, just a method of estimating one. Thanks!

Context:

I'm trying to create a nonlinear slider for the UI in a program I'm writing. I want to be able to specify the logarithmic scale by specifying the what the output value should be at the midpoint. If I map all the values into a (0,1) range, I need to find an invertible function $f(y)$ such that $0 < \tau < 1$, $f(0) = 0$, $f(1) = 1$, $f(\frac{1}{2}) = \tau$, and $f'(y) > 0$ for all $0 < y < 1$, which also has a nice curve so it doesn't seem jumpy to the user. I realized one such function is:

$$f(y) = \log_\alpha\left(1+y\left(\alpha-1\right)\right)$$

Now I just need to find the base $\alpha$ given the midpoint $\tau$, which is where all this comes in. I'm not 100% sure my reasoning on this is correct, but it seems like a plausible starting point.

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There is no algebraic method to find $x$ but we can approximate it well.

Using the change of base rule for logs, we see $$\log_x \left( \frac{x+1}{2} \right) = \frac{ \ln \left( \frac{x+1}{2} \right)}{\ln x}$$ so if we rearrange, we get the equation $$\tau \ln x - \ln \left( \frac{x+1}{2}\right) =0.$$

The reason I rewrote it in this because now we have a root finding problem, which is well studied. In particular, we want to find a value $x$ such that $f(x)=0$ where $f$ is defined by $f(z) = \tau \ln z - \ln \left( \frac{z+1}{2}\right).$ There are many ways to find the roots of a function, the simplest examples being the Bisection Method and the Newton-Raphson method.

For this particular problem, $f$ is quite well behaved and has a simple derivative, so trying Newton-Raphson isn't a bad idea. We compute $f'(x) =\displaystyle \frac{\tau}{x}-\frac{1}{x+1}.$

So Newtwon-Raphson iteration gives $$x_{n+1}= x_n - \frac{ \tau \ln x_n - \ln \left( \frac{x_n+1}{2}\right)}{\frac{\tau}{x_n}-\frac{1}{x_n+1}} .$$

As for an initial estimate, for $0<\tau<1,$ $x_0=2$ isn't a bad start. So once you specify $\tau$ you can start approximating $x.$

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Thanks for your answer! You sort of lost me at the second part, but I'll see what Google turns up for "Newtwon-Raphson iteration" :-) –  OverMachoGrande Jun 13 '12 at 13:23
@Fraser What accuracy do you require $x$ to? If you only need around 4 or 5 decimal places then I'd recommend the Bisection method - it is more stable and easier to implement/understand. –  Ragib Zaman Jun 13 '12 at 13:24
@Fraser Also, in practice is your $\tau$ going to be (hopefully) between say 0.1 and 0.9? Or is it essential that $\tau$ can be quite close to 0 or 1? Because then this problem becomes difficult, the value of $x$ which solves the problem becomes huge if $\tau$ is near $1$ (eg if $\tau=0.98, x\approx 10^{15}$) and extremely small if $\tau$ is near 0 (eg $\tau=0.01, x\approx 10^{-31}$). So computers will find it very hard to find $x$ to any decent accuracy. Perhaps another choice for your function or another approach would be more viable. –  Ragib Zaman Jun 13 '12 at 13:31
I tried implementing the equation you gave and it always goes to 1, usually in a few iterations. I'll check if everything is going right. Yes, the $\tau$ will usually be between 0.1 and 0.9, at least for now. –  OverMachoGrande Jun 13 '12 at 13:50
@Fraser If problems continue to occur with the Newton iteration, just try bisection. –  Ragib Zaman Jun 13 '12 at 14:09