Show that the function $\log x$ cannot be expressed in the form $f(x)/g(x)$, where $f(x)$ and $g(x)$ are polynomials with real coefficients.
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10Don't tell people what to do. Also, at least pretend you're interested in doing at least some of your own work. – Tac-Tics Aug 01 '17 at 13:22
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Any thoughts? Presumably you mean $\log |x|$ yes? – lulu Aug 01 '17 at 13:34
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What tool can I use? – Guillemus Callelus Aug 01 '17 at 13:36
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Try to give some insight about your thoughts or approach to the question... – Aug 01 '17 at 13:49
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More generally one can show that $\log x$ is not an algebraic function. See this answer https://math.stackexchange.com/a/1820789/72031 – Paramanand Singh Aug 01 '17 at 14:53
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Suppose $\log(x)=\frac{f(x)}{g(x)}$ for polynomials $f, g$.
Because $\log(x)$ isn't bounded for $x\to \infty$, we must have $\deg(f)>\deg(g)$.
But we have $$\log'(x)=\frac{1}{x}=\frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}$$ and so
$$g(x)^2=xf'(x)g(x)-xg'(x)f(x)$$
and we get the $2\deg(g)=\deg(xf'(x)g(x)-xg'(x)f(x))$.
But $\deg(xf'(x)g(x)-xg'(x)f(x))\leq \deg(g)+\deg(f)$ and so to get no contradiction with $\deg(f)>\deg(g)$, the leading coefficient of $f'g$ and $g'f$ must be the same. But the leading coefficients are the leading coefficients of $fg$ multiplied with $\deg(f)$ in the first case and $\deg(g)$ in the second. And so we also get a contradiction namely $\deg(f)=\deg(g)$.