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I've got the task to find a logarithm function which contains the following points:

$$\begin{align*} A&(5 \mid 4)\\ B&(3\mid6)\\ C&(2\mid8.5) \end{align*} $$

Now I need to find the logarithm function. My idea was to use $f(x)=a\cdot \log(x-b)+c$ and a system of equations:

$$ \begin{align*} 4&=a\cdot \log(5-b)+c\\ 6&=a\cdot \log(3-b)+c\\ 8.5&=a\cdot \log(2-b)+c\\ \end{align*} $$

I tried to solve for $a$, $b$ and $c$ without success. Does someone have a hint for me?

Michael Hardy
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Sam
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  • I think $a\cdot\log(b-x)+c$ might serve better since the terms increase in opposite directions... – abiessu Dec 03 '13 at 18:45

3 Answers3

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$$\begin{align}\frac{4-c}{a}&=\log(5-b)\\\frac{6-c}{a}&=\log(3-b)\\\frac{8.5-c}{a}&=\log(2-b)\end{align}$$ So we obtain $$ b=5-e^{\frac{4-c}a}=3-e^\frac{6-c}a=2-e^{\frac{8.5-c}a}\\ \begin{align}2&=e^{\frac{4-c}a}-e^{\frac{6-c}a}\\ &=e^{\frac{4-c}{a}}\left(1-e^{\frac2a}\right)\\\log 2&=\frac{4-c}a+\log\left(1-e^{\frac2a}\right)\\c&=4+a\left(\log\left(1-e^\frac2a\right)-\log 2\right)\end{align} $$

Similarly, you may obtain $c=4+a\left(\log\left(1-e^{\frac{4.5}a}\right)-\log3\right)$

So we have $$\begin{align} \log\left(1-e^\frac2a\right)-\log 2&=\log\left(1-e^{\frac{4.5}a}\right)-\log 3\\ \frac12\left(1-e^\frac2a\right)&=\frac13\left(1-e^\frac{4.5}a\right)\\ \frac16=\frac12e^\frac2a-\frac13e^\frac{4.5}a \end{align}$$

A Computer Algebra System will tell you $a\approx-2.41377$. From there you can deduce $b$ and $c$

Tim Ratigan
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  • Thank you. I've found a≈-5.558, b≈1.45 and c≈7.058 in a similar manner. – Sam Dec 03 '13 at 19:21
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    I highly recommend my other answer. It's a lot better. (I didn't edit this one because the approach is still valid, it's just worse) – Tim Ratigan Dec 03 '13 at 19:26
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Found a way easier and more effective way (facepalm)

By subtracting equations, we find $$\begin{align}2&=a\log\left(\frac{3-b}{5-b}\right)\\4.5&=a\log\left(\frac{2-b}{5-b}\right)\\\hline 2.25&=\frac{\log\left(\frac{2-b}{5-b}\right)}{\log\left(\frac{3-b}{5-b}\right)}\\ \left(\frac{3-b}{5-b}\right)^{9/4}&=\frac{2-b}{5-b}\\ (3-b)^{9/4}&=(2-b)(5-b)^{5/4}\\ (3-b)^9&=(2-b)^4(5-b)^5\end{align}$$

This makes $b$ the root of an $8^{th}$ degree polynomial: $$ 0=30317 - 90951 b + 116268 b^2 - 82764 b^3 + 35907 b^4 - 9735 b^5 + 1614 b^6 - 150 b^7 + 6 b^8 $$

Now that you have one variable ($b\approx 1.44969$), it should be simple to solve the rest (that's not to say there will be nice ways to represent $a$ and $c$)

Tim Ratigan
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Hint: is there an exponential function $x\mapsto ab^x$ through $(4,5)$, $(6,3)$ and $(8.5,2)$?

Michael Hoppe
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