Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here's the graph. When I use the points $(-1,1)$ or $(-3,2)$ to use in the equation $a\log(-x-1)+k$, I can't find a finite value for k. Any ideas?

share|cite|improve this question
It should probably be $-x+1$ in the argument of $\log$. When $x=0$, you want to take $\log 1$, not $\log(-1)$. – Rahul Aug 27 '12 at 7:35
Appreciate your help but the vertical asymptote is x=1 so shouldn't the log argument be (x-1)? Since the graph is going away from 0, it would be (-x-1), no? – Mark Aug 27 '12 at 7:39
up vote 0 down vote accepted

It looks like a mirror image of $\log(x)$ around $x=\frac 12$ with a vertical stretching so that the equation could be : $$\alpha \log(1-x)$$ Since $\alpha \log(1-(-3))=2$ I would say that $\alpha=\frac 2{\log(4)}$ with the final : $$\frac {\log(1-x)}{\log(2)}$$ corresponding to the picture :


share|cite|improve this answer
I understand how you got to alog(2) but I'm not so clear how you got to a = 2/(log(4)) since there is a k variable in the equation above. – Mark Aug 27 '12 at 7:45
@Mark: for $x=-3$ we observe $y=2$ so that replacing in $y=\alpha\log(1-x)$ I get $2=\alpha\log(4)=\alpha\log(2^2)=\alpha\ 2\log(2)$ and dividing by $2\log(2)$ we get $\alpha$. – Raymond Manzoni Aug 27 '12 at 7:50
@Mark : I didn't consider $a\cdot \log(-x-1)+k$ since it is clearly wrong ($\log(-x-1)$ should be $\log(1-x)$ and if $k$ is not a constraint but a parameter to find it is simply $0$). – Raymond Manzoni Aug 27 '12 at 7:54
Oh I see, so to find k you just make it zero in order to find a first. Then sub a along with another point to find k to find the log function altogether. – Mark Aug 27 '12 at 7:57
@Mark: if $y=a \log(1-x)+k$ then $k$ is the vertical value for $x=0$ but it is clearly $0$ on the picture so that $k=0$ (you may find it first). After that you consider the other value $x=-3$ getting $a=\frac 1{\log(2)}$ and finally you observe that it works too with $x=-1$. – Raymond Manzoni Aug 27 '12 at 8:01

You're solving for two parameters with two linear equations. Check it out: $$ y_1 = a \log(-x_1+1) + k $$ $$ y_2 = a \log(-x_2+1) + k $$ So solve for $a,k$ as though all other variables are constant: $$ y_1-y_2 = a(\log(-x_1+1) - \log(-x_2+1)) = a \log \frac{-x_1+1}{-x_2+1}~~, $$ and we find $ a = \dfrac{y_1-y_2}{\log \frac{-x_1+1}{-x_2+1}} $ . Plugging in $a$ into either initial equation will yield $k$ .

In our particular example, we can use $(x_1,y_1) = (-1,1), (x_2,y_2) = (-3,2)$ to find that $a = \dfrac{-1}{\log \frac{1}{2}} = \dfrac{1}{\log 2}$ and so we find $k$ from the fact that $$ 1 = \frac{\log2}{\log2}+k = 1+k~~, $$ and so $k = 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.