identity of prime counting function

is this identity trivial ?

$$ \pi (x) = \sum_{n=2}^{x} \frac{\Lambda (n) \mu (n)}{log(n)} $$

the mangoldt function is only nonzero for primes and prime powers and for any power of prime higher that 1 $ \mu(p^m) $

that'

Yes, simply because Mangoldt lambda is non zero function when $n$ is a prime or a prime-power, we can say that $ \sum \limits_{n=2}^{x} \frac{\Lambda(n) \mu(n)}{\ln n} = \sum \limits_{2\leq n\leq x, n= p^{\alpha}} \frac{\Lambda(n) \mu(n)}{\ln n}$

And since Moebius mu is non zero function when $n$ is square-free number, and so $2\leq n \leq x $ that satisfy $n$ is prime or a prime-power and $n$ is square-free numbers are the primes so $ \sum \limits_{n=2}^{x} \frac{\Lambda(n) \mu(n)}{\ln n} = \sum \limits_{ 2\leq p \leq x} \frac{\Lambda(p) \mu(p)}{\ln p}$

$ \Lambda(p) = \ln p $ and $\mu(p) = -1 $ so $ \sum \limits_{ 2\leq p \leq x} \frac{\ln p * (-1)}{\ln p} = \sum \limits_{ 2\leq p \leq x} -1 = -\pi(x) $

So the summation should be with negative sign to be correct