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thanks for clicking on my topic. I have only just finished high school so apologies if this is a stupid question in advance!

I was messing around with the equation $$x^a - b^x = 0$$

I didn't find this from anywhere, was just thinking about it as a further generalisation of $$x^n - n^x = 0$$

So I did the following: $$x^a - b^x = 0$$

$$x^a = b^x$$

$$\ln(x^a) = \ln(b^x)$$

$$a \cdot\ln(x) = x\cdot \ln(b)$$

$$x/\ln(x) = a/\ln (b)$$

The left-hand side above is an asymptotic approximation for the number of primes up to $x$, by the prime number theorem. So by substituting $a$ and $b$ into the right-hand side for a given case, a range for $x$ could be determined based on how many primes precede it. This is obviously useless because computers can solve this already, but I thought maybe it shows some possible link between the prime numbers and exponential and polynomial function intersections. It probably means nothing but I wanted to post this anyway, just to see what others think. Thanks!

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$\dfrac{x}{lnx}$ is asymptotic approximation. And it approaches to the prime counting function $\pi (x)$ as $x$ increases boundlessly but it doesn't give any bound to the error when $x$ is bounded. Therefore, you can't bound the approximation error of $x$ if you use prime number theorem.

Instead, for a given case, since you know the value of $\dfrac{\ln x}{x}$ and its monotonic increasing function. You can find $x$ more easily.

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    $\begingroup$ Actually, it's $x/\log x$ that is asymptotic to $\pi(x)$. And there are refined versions of the Prime Number Theorem that give error terms. $\endgroup$ – Gerry Myerson Dec 13 '16 at 11:54

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