# Can $\ln$ be written as a ratio of polynomials?

Is it possible that $$\ln(x)=\frac{p(x)}{q(x)}$$ for all $$x>0,$$ where $$p$$ and $$q$$ are polynomials with real coefficients?

I think the answer is no. Suppose two such polynomials did exist. Take the limit as $$x$$ goes to infinity. This gives that $$\deg(p)>\deg(q),$$ as $$\lim_{x\to\infty}\ln(x)=\infty.$$ Let $$m=\deg(p)$$ and $$n=\deg(q).$$ Differentiate both sides to get $$\frac{1}{x}=\frac{p'(x)q(x)-p(x)q'(x)}{q(x)^2}.$$

Rearrange (this step is valid as $$x>0$$ and $$q(x)^2>0$$ by hypothesis) to get $$q(x)^2=xp'(x)q(x)-xp(x)q'(x).$$

Let the leading coefficient of $$p$$ be $$a$$ and the leading coefficient of $$q$$ be b. The leading coefficient of $$xp'(x)q(x)$$ then, is $$amb.$$ Similarly, the leading coefficient of $$-xp(x)q'(x)$$ is $$-bna.$$ Suppose their sum were $$0.$$ Then, $$ab(m-n)=0.$$ But, $$ab≠0$$ (as $$a$$ and $$b$$ are leading coefficients). So, $$m-n=0.$$ This contradicts $$m>n.$$ Hence, the coefficient of $$x^{m+n}$$ in the RHS is non-zero. Now, compare degrees to get $$2n=m+n.$$ This contradicts $$m>n.$$

Is my approach right? What other methods can we use to show this?

• This seems right! Alternatively, consider the limit of both sides as you did, then divide both sides by $x$ and consider the new limits. Dec 6, 2023 at 7:24
• You can also use growth rate arguments (which would be similar to Greg's idea) Dec 6, 2023 at 7:29
• Why is the degree of the numerator exactly $m+n-1$? Adding two polynomials of the same degree may result in a polynomial of a lower degree.
– lhf
Dec 6, 2023 at 11:27
• @lhf, yes I missed that. See my new edit; it should have hopefully fixed the error. Dec 6, 2023 at 11:39
• Your proof is correct now, as far as I can tell. – As indicated in the other comments, you can also use that the logarithm does not have “polynomial growth,” i.e. there is no integer exponent $k$ such that $\ln(x) \sim x^k$ for $x \to \infty$. Dec 7, 2023 at 8:21

In this answer no derivatives nor growth rate of $$\ln x$$ are used.
If $$\ln x={p(x)\over q(x)}$$ then $$\deg p>\deg q$$ as the limit at $$\infty$$ is equal $$\infty.$$ We get $${p(x)\over q(x)}=\ln x={1\over 2}\ln (x^2)={p(x^2)\over 2 q(x^2)}$$ Thus $$2p(x)q(x^2)=p(x^2)q(x).$$ Hence $$\deg p+2\deg q=2\deg p+\deg q$$ This implies $$\deg p=\deg q,$$ a contradiction.