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I have a presentation in which I want to point out a difference between two functions. Instead of putting the two functions on a slide and pointing to the differences, I want to do something simple.

The first function has a form of $f_1(x) = g(x)\log (h(x))$ and the second function has the form $f_2(x) = g(x)\log(r(x)h(x))$.

The thing is that saying something along the lines "$f_2(x) - f_1(x)$ have the form of $g(x)\log r(x)$ keeps the $g(x)$ (which is rather complicated) and saying something like $f_2(x)/f_1(x)$ has the form $\log h(x) / \log (r(x)h(x))$ is also cumbersome.

Does anyone see a clear way to point out the difference between these two functions by only making explicit the additional $r(x)$?

I would really like to make it simple, ideas are welcome.

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    $\begingroup$ How about $f_1/g - f_2/g$? $\endgroup$ – Aryabhata Jan 2 '11 at 3:39
  • $\begingroup$ How about $\frac{f_2}{f_1}=1+ \log _{r(x)} h(x) $? $\endgroup$ – Arjang Jan 2 '11 at 5:03
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The short answer to your question is no.

To see this, define $a(x) = g(x) \log(h(x))$.

So we have $a(x) = f_1(x)$ and $a(x) + \log(r(x)) g(x) = f_2(x)$.

So here we have two linearly independent equations in $a(x)$ and $g(x)$.

Hence at best we can solve for $a(x)$ and $g(x)$ (i.e. we can solve for $h(x)$ and $g(x)$) in terms of $f_1(x)$ and $f_2(x)$.

So, either we need another equation, so that we can plug in what we obtain to get rid of $g(x)$ and $h(x)$ or We need $g(x)$ or $h(x)$ to be some contant so that the dependence of a relation between $f_1(x),f_2(x)$ and $r(x)$ on $g(x)$ or $h(x)$ is hidden.

You could however define new functions $\tilde{f_1}$ and $\tilde{f_2}$ depending on $f_1(x)$,$f_2(x)$,$g(x)$ and $h(x)$ and massage to get a relation between $\tilde{f_1}$ and $\tilde{f_2}$.

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instead of $f_1,f_2$ would you rather consider $e^{f_2-f_1}={r(x)^{g(x)}}$?

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