How do you prove that for all real numbers c, there always exists a value n, in which $4^n > c3^n$?

Question

What we have so far, its $\frac{4^n}{ 3^n} > c,$ such that $\left(\frac{4}{3}\right) ^ n > c.$ From that I know that it is true, but I don't really know how to prove it formally.

Answer 1

For non-positive $c$'s this is trivial. So let $c>0$. Then, we know by the Archimedean property that there is some $n_0 \in \mathbb N$ such that:

$$\frac1{n_0} < \frac{\log(4/3)}{\log c}$$

Answer 2

Hint.

If $c\leq0$, then the result is clear.

If $c>0$, then you may take $$ n>\frac{\ln c}{\ln(4/3)}. $$

Answer 3

Note that if $n$ is a positive integer then $$\left(\frac{4}{3}\right)^n =\left(1+\frac{1}{3}\right)^n\ge 1+\frac{n}{3}$$ (Bernoulli Inequality).

Answer 4

We want $$4^{n} > c3^{n} = \frac{4^n}{3^n} > c = (\frac{4}{3})^n > c $$

By using the relationship between exponential functions and log, we can get $$ n > log_\frac{4}{3} c$$

So now, for any c that you pick, you can always find n such that $$ n > log_\frac{4}{3} c$$ which will satisfy the original equation.

QED