Showing $\pi(ax)/\pi(bx) \sim a/b$ as $x \to \infty$

I'm having a bit of a problem with exercise 4.12 in Apostol's "Introduction to Analytic Number Theory". I don't think it's supposed to be a very hard exercise, it's the first one in its section (they're usually a bit like warm-ups). I'm supposed to show that

If $a>0$ and $b>0$, then $\pi(ax)/\pi(bx) \sim a/b$ as $x \to \infty$.

It also says I'm allowed to use the prime number theorem. Is it just something like (a rough sketch): $$\frac{\pi(ax)}{\pi(bx)} \sim \frac{ax \log bx}{bx \log ax} \sim \frac{a}{b}, \quad \text{since the logs \to 1 as x \to \infty?}$$ I don't know, maybe I'm heading in the wrong direction... It would be very nice if someone could show me how to do this properly!

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Looks good to me, although instead of saying "the logs go to 1" I would say the ratio of the logs goes to 1. –  Gerry Myerson Dec 2 '11 at 9:53
@Gerry: Oh, ok, I thought I was a long way from being done! Thanks! –  Carolus Dec 2 '11 at 10:13

All you're missing is the use of $\text{log}(bx) = \text{log}(b) + \text{log}(x)$, as the fraction then ~$\frac{ax\text{log}(x)}{bx\text{log}(x)}$ ~ $\frac{a}{b}$.

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You have $\frac{ax \log bx}{bx \log ax} = \frac ab \frac{\log x+\log b}{\log x+\log a}$.

Since $\log x\to\infty$, you have $$\lim\limits_{x\to\infty} \frac{\log x+\log b}{\log x+\log a}=1.$$ (The constants $\log a$ and $\log b$ are "small" compared to $\log x$.)

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Of course! Thank you! –  Carolus Dec 2 '11 at 10:14

It might be interesting to note that the initial statement is what guarantees that the fractions of the form $p/q$ with $p,q$ primes, form a dense subset of $[0,\infty).$

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