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As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let $$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{\nu_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem III.5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book "Introduction to analytic and probabilistic number theory". This section is contained in pages 516-517.

If we let $$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{\nu_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper "On numbers with small prime divisors" (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let $$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{a_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let $$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{\nu_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let $$N_k(z):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le z\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{z^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(z)\le \frac{(z+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$.

  This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.

For a reference see Theorem 5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book ``Introduction to analytic and probabilistic number theory''. This section is contained in pages 516-517.

If we let $$N_k(x):=\left| \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right|$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{a_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$-friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper ``On numbers with small prime divisors'' (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.