Can you use a logarithm coefficient in a linear equation?

I have an equation that looks like $x+(\ln3)y+z=0$ where there's a natural logarithm as a coefficient. Is it possible to have this in a linear equation? I know that you cannot have a root or a product of variables in a linear equation, but I'm not so sure about coefficients that include an exponent.

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$\ln 3$ is just a number, same as $5$ or $\pi$ or $\sqrt 2$, so yes. If you had $\ln x$ in there it would be a problem. – Rahul Sep 25 '11 at 21:35

Yes, you can. You should distinguish parameters/coefficients and variables.

1. This is a linear equation: $$x+(\log 3)y +z = 0$$

2. This is a non-linear equation: $$x+\log (3y)+z = 0$$

3. This is a linear equation for $x,y$: $$x+(\log k)y = 0$$ and non-linear if variables are $x,y$ and $k$.

Simple test which is in fact a definition of a linear equation is the following. Let you have an equation $$f(x,y,z) = 0$$ e.g. in your case $f(x,y,z) = x+(\log 3)y+z$. To check it's linearity it's necessary and sufficient to have: $$f(\alpha x'+\beta x'',\alpha y'+\beta y'',\alpha z'+\beta z'') = \alpha f(x',y',z')+\beta f(x'',y'',z'').$$

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You may want to be a bit careful with "linearity". For example, it is not uncommon in the context of calculus to refer to a function of the form $f(x)=mx+b$ as a "linear function", even though it is not linear (in the sense of linear algebra) unless $b=0$... – Arturo Magidin Sep 25 '11 at 21:56
You're right, though I thought it's more usual to call them affine. – Ilya Sep 25 '11 at 21:59
Alas, not in calculus books that I am familiar with, though absolutely in linear algebra books. – Arturo Magidin Sep 25 '11 at 22:01
@ArturoMagidin: I wouldn't argue since I've studied calculus in another language and started to deal with linear functions (in English) only through the linear algebra and theory of linear operators (that's why I don't like affine function to be called linear). Btw, what is Alas? – Ilya Sep 25 '11 at 22:05
"Alas" is (more or less) a poetic way of saying "Unfortunately". – Henning Makholm Sep 25 '11 at 22:21