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I am given two points $(1.5, 3.1)$ and $(6.5, 1.45)$ and I have to find a function that cuts through these points. I have chosen to use an $\ln(x)$ function in the general form:

$y=a\ln(x-h)-k$

I know that an additional point is needed to find the values of $a$, $h$ and $k$ however i want to write $h$ and $k$ in terms of $a$. Then i can simply test different values of $a$ to see which one fits my needs.

paulplusx
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Deep Patel
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    Have you tried plugging in your two points? You get two equations in three unknowns. Now try to solve them for $h$ and $k$, treating $a$ as known. That gives two equations in two unknowns. I don't know if the $\ln$ will be a problem. Please try and show what you have. – Ross Millikan Sep 15 '18 at 05:03

1 Answers1

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When you have the value of $h$ finding the value for $k$ is easy:

$k=a\ln(x_1-h)-y_1$

So the question is: How to find $h$ when $a$ is given?

You have two equations:

$y_1=a\ln(x_1-h)-k\\ y_2=a\ln(x_2-h)-k$

($(x_1,y_1)$ and $(x_2,y_2)$ are the two given points.)

Subtract the two equations from each other, divide both sides by $a$ and you'll get:

$\frac{y_1-y_2}a=\ln(x_1-h)-\ln(x_2-h)$

You may now exponentiate both sides of the equation:

$e^{\frac{y_1-y_2}a}=\frac{x_1-h}{x_2-h}$

... which will lead to a rather complex term for $h$ which is depending on $a$.

If I did not make a mistake it is:

$h = \frac{x_1-x_2e^{\frac{y_1-y_2}a}}{1-e^{\frac{y_1-y_2}a}}$

Then you can insert that term into the term $k=a\ln(x_1-h)-y_1$ ...

Note that not all values of $\mathbb R$ are allowed for $a$!

Martin Rosenau
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