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Graphing y = log(x)/log(2x-1)

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The graph:

from to

Intersection points:

does show?

Piecewise:

The solution

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          log(x)   
f(x) = ------------
       log(2*x - 1)
$$f{\left(x \right)} = \frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}}$$
f = log(x)/log(2*x - 1)
The graph of the function
The domain of the function
The points at which the function is not precisely defined:
$$x_{1} = 1$$
The points of intersection with the X-axis coordinate
Graph of the function intersects the axis X at f = 0
so we need to solve the equation:
$$\frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}} = 0$$
Solve this equation
Solution is not found,
it's possible that the graph doesn't intersect the axis X
The points of intersection with the Y axis coordinate
The graph crosses Y axis when x equals 0:
substitute x = 0 to log(x)/log(2*x - 1).
$$\frac{\log{\left(0 \right)}}{\log{\left(-1 + 0 \cdot 2 \right)}}$$
The result:
$$f{\left(0 \right)} = \tilde{\infty}$$
sof doesn't intersect Y
Extrema of the function
In order to find the extrema, we need to solve the equation
$$\frac{d}{d x} f{\left(x \right)} = 0$$
(the derivative equals zero),
and the roots of this equation are the extrema of this function:
$$\frac{d}{d x} f{\left(x \right)} = $$
the first derivative
$$- \frac{2 \log{\left(x \right)}}{\left(2 x - 1\right) \log{\left(2 x - 1 \right)}^{2}} + \frac{1}{x \log{\left(2 x - 1 \right)}} = 0$$
Solve this equation
Solutions are not found,
function may have no extrema
Inflection points
Let's find the inflection points, we'll need to solve the equation for this
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = 0$$
(the second derivative equals zero),
the roots of this equation will be the inflection points for the specified function graph:
$$\frac{d^{2}}{d x^{2}} f{\left(x \right)} = $$
the second derivative
$$\frac{\frac{4 \left(1 + \frac{2}{\log{\left(2 x - 1 \right)}}\right) \log{\left(x \right)}}{\left(2 x - 1\right)^{2} \log{\left(2 x - 1 \right)}} - \frac{4}{x \left(2 x - 1\right) \log{\left(2 x - 1 \right)}} - \frac{1}{x^{2}}}{\log{\left(2 x - 1 \right)}} = 0$$
Solve this equation
The roots of this equation
$$x_{1} = 70130.7624851036$$
$$x_{2} = 83773.4524097231$$
$$x_{3} = 168591.178470941$$
$$x_{4} = 111554.643610193$$
$$x_{5} = 163777.3003064$$
$$x_{6} = 88360.8619389959$$
$$x_{7} = 217219.586169279$$
$$x_{8} = 139862.835233527$$
$$x_{9} = 158973.334359218$$
$$x_{10} = 212319.984842555$$
$$x_{11} = 106884.23209112$$
$$x_{12} = 178247.556996465$$
$$x_{13} = 116239.554226526$$
$$x_{14} = 79205.2297330639$$
$$x_{15} = 154179.579113005$$
$$x_{16} = 183089.53622477$$
$$x_{17} = 135113.283271347$$
$$x_{18} = 202543.894024119$$
$$x_{19} = 149396.350599736$$
$$x_{20} = 102228.942081859$$
$$x_{21} = 120938.388540661$$
$$x_{22} = 97589.4489170793$$
$$x_{23} = 74657.2680720997$$
$$x_{24} = 187940.383254848$$
$$x_{25} = 192799.868958311$$
$$x_{26} = 92966.4877886856$$
$$x_{27} = 130375.732035869$$
$$x_{28} = 197667.775189082$$
$$x_{29} = 207428.027072087$$
$$x_{30} = 144623.983966592$$
$$x_{31} = 125650.613100783$$
$$x_{32} = 227041.033321087$$
$$x_{33} = 222126.657681893$$
$$x_{34} = 173414.686516363$$
You also need to calculate the limits of y '' for arguments seeking to indeterminate points of a function:
Points where there is an indetermination:
$$x_{1} = 1$$

$$\lim_{x \to 1^-}\left(\frac{\frac{4 \left(1 + \frac{2}{\log{\left(2 x - 1 \right)}}\right) \log{\left(x \right)}}{\left(2 x - 1\right)^{2} \log{\left(2 x - 1 \right)}} - \frac{4}{x \left(2 x - 1\right) \log{\left(2 x - 1 \right)}} - \frac{1}{x^{2}}}{\log{\left(2 x - 1 \right)}}\right) = -0.500000000000001$$
$$\lim_{x \to 1^+}\left(\frac{\frac{4 \left(1 + \frac{2}{\log{\left(2 x - 1 \right)}}\right) \log{\left(x \right)}}{\left(2 x - 1\right)^{2} \log{\left(2 x - 1 \right)}} - \frac{4}{x \left(2 x - 1\right) \log{\left(2 x - 1 \right)}} - \frac{1}{x^{2}}}{\log{\left(2 x - 1 \right)}}\right) = -0.5$$
- the limits are not equal, so
$$x_{1} = 1$$
- is an inflection point

Сonvexity and concavity intervals:
Let’s find the intervals where the function is convex or concave, for this look at the behaviour of the function at the inflection points:
Have no bends at the whole real axis
Vertical asymptotes
Have:
$$x_{1} = 1$$
Horizontal asymptotes
Let’s find horizontal asymptotes with help of the limits of this function at x->+oo and x->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the left:
$$y = 1$$
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}}\right) = 1$$
Let's take the limit
so,
equation of the horizontal asymptote on the right:
$$y = 1$$
Inclined asymptotes
Inclined asymptote can be found by calculating the limit of log(x)/log(2*x - 1), divided by x at x->+oo and x ->-oo
$$\lim_{x \to -\infty}\left(\frac{\log{\left(x \right)}}{x \log{\left(2 x - 1 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the right
$$\lim_{x \to \infty}\left(\frac{\log{\left(x \right)}}{x \log{\left(2 x - 1 \right)}}\right) = 0$$
Let's take the limit
so,
inclined coincides with the horizontal asymptote on the left
Even and odd functions
Let's check, whether the function even or odd by using relations f = f(-x) и f = -f(-x).
So, check:
$$\frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}} = \frac{\log{\left(- x \right)}}{\log{\left(- 2 x - 1 \right)}}$$
- No
$$\frac{\log{\left(x \right)}}{\log{\left(2 x - 1 \right)}} = - \frac{\log{\left(- x \right)}}{\log{\left(- 2 x - 1 \right)}}$$
- No
so, the function
not is
neither even, nor odd