How to figure out the starting point for this Mandelbrot? My answer for another math stackexchange question, asked by Gottfried, involved observing Mandelbrot bifurcation for the iterated function in question, $f(z)\mapsto z-\log_b(z)$.  In particular, for each base, there's a well defined Julia, and if $\Re(z)<-1$, then we can say the iterates of f(z) get arbitrarily large negative.  For other points we observe f(z) iterates towards a stable attracting cyclic orbit.  The first bifurcation occurs at $b=\exp(0.5)$.  For $\Re(b)>\exp(0.5)$, z=1 is an attracting fixed point, and for For $\Re(b)<\exp(0.5)$, z=1 is a repelling fixed point.
Then I wondered whether a Mandelbrot plot could be generated as the logarithmic base b varies.  I got as far as generating a Mandelbrot plot for the escape iteration count, iterating of f(z), but I'm having some serious difficulties.  I'm still puzzled as to the correct algorithm to generate these Mandelbrots, because the required starting point seems to be a function of b, the logarithmic base.  Is there an algorithm to calculate a good starting point for iterating the Mandelbrot $f(z)\mapsto z-\log_b(z)$, such that the starting point is guaranteed to be in the cyclic attracting basin for b, if base(b) has an attracting cycle?  
For the normal Mandelbrot, $f(z)\mapsto z^2+c$, the starting point used for each value of "c" is z=0, which is the center of the corresponding Julia for c.  The exterior of the Julia can be put into correspondence with a Botcher function for z^2, for $|z|>1$.  Bonus question: Is there a corresponding Botcher function for Julias for Gottfried's f(z) function, and can these Julias be put into correspondence with the Julia's for the normal Mandelbrot set?  Gottfried's Julia's are not symmetrical, and actually have infinitely large positive points that iterate into the stable attracting basin.  
For example, there are flaws in this Mandelbrot plot, where some of the points that are colored really belong to a cyclic basin, and instead should be black since the cyclic basin never escapes.  Here, I started iterating with z=2.6, which is an ok starting point for this plot, but its not perfect. I tried other plots, where I use multiple starting points, which is a little better, but far from ideal.  This plot varies from b=1.425 to b=1.725 with grid lines of 1/10th. 

 A: As noted by Sheldon, the good starting point must be a critical point. There is indeed a theorem that says that if you have an attracting cycle, then at least one critical point must belong to its attraction basin. 
Roughly, the idea of the proof is as follows : around an attracting fixed point, there is a linearizing coordinate $\phi$ such that $f(\phi(z))=\lambda \phi(z)$. That coordinate is defined only in a neighborhood of the attracting fixed point, however using the functional equation it satisfies it is possible to prolonge it until you meet a critical point.
Now if you have a cycle instead of a fixed point, just replace $f$ by $f^p$ to get the same result.
The Mandelbrot set is of mathematical interest because in complex dynamics, the global behaviour of the dynamics is generally ruled by the dynamics of the critical points. Thus knowing the dynamics of the critical points give you information on all of the dynamics. In the simplest family $z^2+c$, there is only 1 critical point (0) and so it is natural to look at what happens to its dynamics depending on $c$. 
If you want to generalize the notion of Mandelbrot set for, say, cubic polynomials $z^3 + az+b$, you would have to look at the behaviour of two critical points, and so not only would you get a set in $\mathbb{C}^2$, you would also need to make a choice in your definition : are you looking at parameters where both critical points are attracted to a cycle, or one of them, or none ?
In your case, there is only one critical point, so your set is a reasonable analogue of the Mandelbrot set.
EDIT : note that the definition of the Mandelbrot set does not use attracting cycles, but depends on whether or not the critical point goes to infinity. It is conjectured (it's one of the most important conjecture in the field) that the interior of the Mandelbrot set is exactly composed of parameters for which the critical point is attracted to a cycle. However it is well known that in the boundary of the Mandelbrot set, you have no attracting cycles.
EDIT 2 : One of the most interesting features of the Mandelbrot set is that its boundary is exactly the locus of bifurcation, i.e. the set of parameters for which the behaviour of the dynamics changes drastically. If you choose any holomorphic family of holomorphic maps $f_\lambda$, you can also define the locus of bifurcation for this family. It has been proved that this set is either empty or contains copies of the Mandelbrot set.
A: This is a partial answer, not a complete answer.  $z_0=\frac{1}{\log(b)}$ is a good starting point $z_0$, such that iterating $f(z)\mapsto z-\log_b(z)$ will converge towards a cyclic fixed point for logarithmic base b, if there is a cyclic fixed point.  The value $z_0$ seems to work perfectly for all values of b tried, and allows generation of beautiful Mandelbrot plots for iterating f(x). At the
chosen value for $z_0$, the derivative of $f(z_0+x)$ is zero.  This is analogous to iterating $z^2+c$ starting at
$z_0=0$, where the derivative is zero.
To derive $z_0$, calculate the zero of the first derivative $\frac{d}{dz}(z-\log_b(z))=1-\frac{1}{z\log(b)}=0$.  Solving for z,
we get $z\log(b)=1$ and then $z_0=\frac{1}{\log(b)}$.
I don't know how to prove it works.  It works for all bases I've tried.  It is useful for generating Mandelbrot pictures for iterating f(z) in the complex plane, including the pictures below.  Once you have such a value for $z_0$, it becomes possible to make accurate Mandelbrot plots, which I couldn't do before.  Naturally one asks what the global picture of the Mandelbrot for Gottfried's function looks like. I did change the escape criteria to $\Re(z)<-4.5$.  This image used 2000 iterations.  The center "black" region is not correct.  For bases close to zero, the iterates of f(x) escape to plus infinity instead of minus infinity.  The grid lines are 0.5 space apart, with the image varying from -1.66 to +1.66.  Notice the distortion in the bug on the cutline at the real axis for negative values of z; the spiral actually continues infinitely.  Right click on the image to enlarge and open in a separate tab.

The second image varies from 1.44 to 1.66, with grid lines of 1/10th.  This image would replace the flawed image I posted last time.  I also have updated blowups from all other previous images I posted to the earlier question, and they are also flawless with the new initial starting point.

You can see from the plot, that the Mandelbrot looks like a giant infinitely spiraling circle, centered at b=0, with the radius of the circle as $\exp(0.5)$, which will be derived below. We already know that the primary 2:1 bifurcation occurs at $\exp(0.5)$, from the previous post on this equation.  The boundary for this circle would be where the primary fixed point switches from being attracting, to being repelling.  On the boundary, the fixed point would be neutral. For f(z), the primary fixed point is always 1. Start by assuming that $b(k)=\exp(0.5+ki)$ is the boundary, where k is a real number, varying from +/- infinity.  This equation for b(k) defines an infinite logarithmic spiral of radius exp(0.5).  Next, we show that on the boundary for this spiral, the fixed point of 1 is indifferent, neither attracting, nor repelling.  The fixed point of 1+delta has a derivative whose absolute value is 1.
Equation to calculate the first derivative at fixed point=1, and then substitute in the proposed solution for the base $b=\exp(0.5+ki)$.
$f(1+z)-1$
$1+z - \log_b(1+z) - 1$
$1+z - (z-z^2/2+z^3/3....)/\log(b) -1 = z - (z-z^2/2+z^3/3....)/\log(b)$
Next, calculate the first derivative at z=0. Observe that for real values of k, s
the |deriv|=1, since the absolute value of the numerator and the denominator are the same.
$(1-1/\log(b)) = \frac{2ki - 1}{2ki+1}$.
With a little more work, this equation can also be used to calculate where any of the primary fixed point n-way bifurcations occur, for any value of n, which I also derived and verified.  For n=2, we get k=0, and we know from previous work that the primary bifurcation point is exp(0.5).
$k =0.5\tan(0.5\pi-\pi/n)$
Note the singularity for n=1.  Iterating f(x) doesn't have a main parabolic cusp of the Mandelbrot set.  One problem I don't have the solution for is I don't know what the correct escape limit equation is.  For the graphs included in this answer, I used $\Re(z)<-4.5$, but perhaps a larger value is required.  Also, for bases close to zero, and less than 1, the function escapes to +infinity instead of minus infinity, so the escape criteria also needs to be fixed for those bases.  I had originally thought about the problem for bases>1 only, and my escape criteria is not accurate for complex bases, especially negative complex bases, and complex bases with real(base)<1.  That's a separate question.
