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?

Solution 1:

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.

Solution 2:

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.

Solution 3:

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....

Solution 4:

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.