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Idk? try the qualified helper thread, it might help ;)

so maybe we can somehow do this by contradiction

\[\log(x)=\frac{f(x)}{g(x) } \text{ where } f \text{ and } g \text{ are polynomials }\]

\[10^{\frac{f(x)}{g(x)}}=x\]

just thinking so far

I think contradiction works well...

just trying to get it to work :p

maybe so

yeah that sounds better, il modify..

like maybe we can do something like that for polynomials of greater degree

on top and bottom

I think we may cross-multiply before differentiatiating..

fine we have a problem at x=0

unless the bottom polynomial is x

or has a factor of x

\[\frac{1}{x \ln(10)}=\frac{f(x)}{xg(x)}\]

probably shouldn't have used f and g

so you think cross multiplying before differentiating could lead to better things maybe?

was thinking.. we might be able to compare coefficients both sides then..

actually that might work solve for log
and compare log expression from before

looks it is better to differentiate first, that way we wont have log in the equation

ahh we could do that too

\[\log(x)=\frac{ \sum_{}^{} i b_i x^i- \sum a_i x^i}{\sum i a_i x^i}\]

g(x) = logx*f(x) satisfies the equation
logx = (g'(x) - f(x)/x)/f'(x)
so i dont see a problem ?

unless f'(x)=c
which means f(x) would have to be xc+d

where c and d are constants

oops

but we need xf'(x) to be a constant so g(x) can be a polynomial

it need not be a constant, we just want \(xf'(x)\) to divide evenly into the numerator, right

yeah I guess it is possibly that the bottom could divide the top

the numerator does have a greater degree

I think we can simply let x > 0

so in other words, we cannot have that \(\log x\) is a rational function

Maybe another rout would be to use the fact that \(e\) is transcendental.