Show that, for $t>0$, $\log t$ is not a polynomial. How can I show that? I've tried to reverse the logarithm to it's exponential form in a trial to show that but I got no success. Can you help me?
 A: Do you mean $t \mapsto \log t$ ? If this is the case, consider the derivatives and the fact that a polynomial's derivatives are eventually null.
A: You can also use the fact that $\lim_{t \to \infty} \frac{ \log t}{t} = 0 \,.$
For a polynomial of degree at least 2, the corresponding limit is infinity, while for a linear polynomial, the limit is zero if and only if the polynomial is constant....
A: What do you mean? Even if the domain is restricted to $(0,\infty)$, a polynomial function will have a finite right hand limit as $t\rightarrow0^+$, while the logarithmic function has not.
A: It does not match in 0: $\lim_{x \to 0^+} \log x = -\infty$ but for any polynomial $\lim_{x \to 0^+} (a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0) = a_0$. Hence there is no $f(x) = (a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0)$ such that $\lim_{x \to 0} (\log x - f(x)) = 0$ so as we get closer to 0 we get divergence from any polynomial.
The $\lim_{x \to 0} \log x$ can be obtained as $\log a = b \Leftrightarrow a = e^b$ and $\lim_{x \to -\infty} e^x = 0$ (assuming logarithm as reverse of exponentiation is well defined).
A: The series $\sum \frac{1}{n \log n}$ diverges and the series $\sum \frac{1}{n (\log n)^2}$ converges, and this isn't true with $\log n$ replaced by any polynomial (with finitely many terms removed if the polynomial has integer zeroes) or indeed with any rational function (and most of the other arguments given so far are not enough to prove this). Actually this argument probably shows that $\log n$ is not even algebraic. 
A: If the logarithm were a polynomial, we could write $$\log(x) = a_n x^n +a_{n-1}x^{n-1}+\cdots +a_0,$$ for constants $a_0,a_1,\ldots,a_n$, where the formula holds for all $x>0$.  Assuming this is the case, a contradiction will be reached.  Using $\log(x^2)=2\log(x)$, we have $$a_n x^{2n}+a_{n-1}x^{2n-2}+\cdots + a_0 = 2a_n x^n+2a_{n-1}x^{n-1}+\cdots+2a_0.$$  Using the fact that equal polynomials have equal coefficients, this implies that $\log(x)=0$ for all $x$. That isn't true, and thus the assumption that $\log$ is a polynomial leads to a contradiction.
A: The function $\log(t)$ can be expressed via Taylor series as the infinite summation
$$
\log(t) = (t-1)-\frac{1}{2}(t-1)^2+\frac{1}{3}(t-1)^3-\frac{1}{4}(t-1)^4+\cdots,
$$
which involves terms of $t^n$ for unbounded $t$.
A: If $\log(t)$ is a polynomial, its derivative $\frac{1}{t}$ also is. But this is clearly impossible (the only invertible polynomials are constants because of the degree).
A: The function $f:t\mapsto \log t$ has a vertical asymptote at $0$ (i.e. $\lim_{t\to0}f'(t)=\infty$), so it has a non-removable discontinuity. Polynomials are continuous everywhere.
