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.
Can you use a logarithm coefficient in a linear equation?
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$\begingroup$
linear-algebra
logarithms
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4$\ln 3$ 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. – 2011-09-25
1 Answers
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Yes, you can. You should distinguish parameters/coefficients and variables.
This is a linear equation: $ x+(\log 3)y +z = 0 $
This is a non-linear equation: $ x+\log (3y)+z = 0 $
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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0@HenningMakholm: thank you. – 2011-09-26