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?
Finding a logarithmic function from a graph
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0Appreciate 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? – 2012-08-27
2 Answers
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 :
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0@Mark: You are welcome ! Fine continuation ! – 2012-08-27
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 $k = 0$.