I know that to plot a straight line that intersects the axis in point $(x_1,y_1)$,(x_2,y_2)$ one can use this equation.

$$(x_2-x_1)\cdot (y-y_1)=(y_2-y_1)·(x-x_1)$$

To be more specific if i want a line that intesects the axis in a given point, I can use:

$$x_1(y-y_2)=- y_2 x$$

but I need to plot a logarithmic function that will intersect the axis in $2$ designated points. for example I know that:

$$y=-\log(0.5+x)$$ intersects $y$ in $0.301$

Can anyone point me to a method of finding such an equation ?

  • $\begingroup$ The logarithm function will always intersect the $x$ axis at $x=1$, and will never touch the $y$ axis... $\endgroup$ – 5xum Jul 7 '16 at 10:51
  • $\begingroup$ what about y=-log(0.5+x) ? it intersects y in 0.301 meaning a valid point on the plot is: (0.301,0) $\endgroup$ – Mortalus Jul 7 '16 at 10:53
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    $\begingroup$ Well that's not the logarithm function, that's a shifted logarithm function. If you tell us all that you are allowed to do with the function, then your question has an answer, of course. So, are you allowed to do $\log(x+A)$? Can you also do $B\cdot \log(x)$? How about $\log(x) + C$? Tell us the rules of the game, and we can tell you how to play it... $\endgroup$ – 5xum Jul 7 '16 at 10:54

Let us go with $y = \alpha \log (x+\beta)$. Let us say that you want the axis intersections to be $(a, 0)$ and $(0, b)$.

Then: $0 = \alpha \log(a + \beta) \Rightarrow a + \beta = 1 \Rightarrow \beta = 1-a $.

Similarly: $b = \alpha \log \beta \Rightarrow \alpha = \frac{b}{\log (1-a)}$.

So, for your logarithmic function to intersect the axes at $(a,0)$ and $(0, b)$ it is enough to take $$y = \frac{b}{\log (1-a)} \log (x + 1 - a)$$

  • $\begingroup$ can you pleas elaborate regarding $0 = \alpha \log(a + \beta) \Rightarrow a + \beta = 1 \Rightarrow \beta = 1-a $ ? $\endgroup$ – Mortalus Jul 7 '16 at 11:33
  • $\begingroup$ Assuming that $\alpha \neq 0$ then we require $\log(a+ \beta) = 0 \Rightarrow a + \beta = e^0 = 1 \Rightarrow \beta = 1- a$ $\endgroup$ – Zain Patel Jul 7 '16 at 11:34

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