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I have equations for two lines, one of which is linear and the other is logarithmic, ie:

$$y = m_1 x + c_1$$

$$y = m_2 \cdot \ln(x) + c_2$$

..and I need to find out where (if at all) these lines intersect. I realise that I need to solve:

$$ m_1 \cdot x + c_1 = m_2 \cdot \ln(x) + c_2$$

..for $x$, but apart from shuffling the constants around I'm not sure how to do this.

Is there a general solution to this problem?

Thanks

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2 Answers 2

up vote 2 down vote accepted

There is no algebraic solution. You can solve this numerically. As log changes so slowly, iterative methods converge quickly.

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Thanks for this. I put together a numeric solution which works fine. It's good to know I wasn't missing an easy way to do this algebraically. –  Simon Andrews Mar 9 '11 at 11:00

The general solution involves the Lambert W function, whose defining equation is $z = W(z)e^{W(z)}$ for complex numbers $z$. If either $m_1$ or $m_2$ is zero in the given problem, then the solution is elementary, so suppose $m_1,m_2 \neq 0$. Then $$ m_1 x + c_1 = m_2 \ln x + c_2 $$ can be rewritten as $$ -\frac{m_1}{m_2}e^{(c_1-c_2)/m_2} = -\frac{m_1 x}{m_2} e^{-m_1 x / m_2}, $$ which has the solution $$ -\frac{m_1 x}{m_2} = W\left(-\frac{m_1}{m_2}e^{(c_1-c_2)/m_2}\right), $$ or $$ x = -\frac{m_2}{m_1} W\left(-\frac{m_1}{m_2}e^{(c_1-c_2)/m_2}\right). $$ For instance, if $m_1=1$, $m_2=-1$, and $c_1=c_2=0$, this gives $x=W(1)=\Omega=0.56714329...$ (the Omega constant), which is correct, since $\Omega = -\ln\Omega$.

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