1. **Stating the problem:** We are given the equation $3^x = 5^y = (75)^z$ and need to show that $xy = z(y + 2x)$. 2. **Rewrite the equation:** Since $3^x = 5^y = (75)^z$, let this common value be $k$. So, $$3^x = k, \quad 5^y = k, \quad (75)^z = k.$$ 3. **Express $k$ in terms of $x$ and $y$:** From the first two, $$k = 3^x = 5^y.$$ 4. **Express $(75)^z$ in terms of prime factors:** Note that $$75 = 3 \times 25 = 3 \times 5^2,$$ so $$(75)^z = (3 \times 5^2)^z = 3^z \times 5^{2z}.$$ 5. **Since $(75)^z = k$, we have:** $$3^z \times 5^{2z} = k.$$ 6. **Recall $k = 3^x = 5^y$, so:** $$3^x = 3^z \times 5^{2z} = 5^y.$$ 7. **Equate the expressions for $k$ in terms of prime bases:** Since $k = 3^x = 5^y$, and also $k = 3^z \times 5^{2z}$, we can write $$3^x = 3^z \times 5^{2z}$$ and $$5^y = 3^z \times 5^{2z}.$$ 8. **Rewrite $k$ as $3^x$ and $5^y$ and equate the logarithms:** Taking logarithm base 3, $$x = z + 2z \log_3 5,$$ and taking logarithm base 5, $$y = z \log_5 3 + 2z.$$ 9. **Use change of base formula:** $$\log_3 5 = \frac{1}{\log_5 3}.$$ 10. **Let $a = \log_5 3$, then $\log_3 5 = \frac{1}{a}$. Substitute:** $$x = z + 2z \times \frac{1}{a} = z + \frac{2z}{a},$$ $$y = z a + 2z.$$ 11. **Calculate $xy$:** $$xy = \left(z + \frac{2z}{a}\right)(z a + 2z) = z^2 \left(1 + \frac{2}{a}\right)(a + 2).$$ 12. **Expand the product:** $$\left(1 + \frac{2}{a}\right)(a + 2) = a + 2 + \frac{2}{a} a + \frac{2}{a} \times 2 = a + 2 + 2 + \frac{4}{a} = a + 4 + \frac{4}{a}.$$ 13. **Rewrite $xy$ as:** $$xy = z^2 \left(a + 4 + \frac{4}{a}\right).$$ 14. **Calculate $z(y + 2x)$:** $$y + 2x = (z a + 2z) + 2 \left(z + \frac{2z}{a}\right) = z a + 2z + 2z + \frac{4z}{a} = z a + 4z + \frac{4z}{a}.$$ 15. **Multiply by $z$:** $$z(y + 2x) = z^2 a + 4 z^2 + \frac{4 z^2}{a} = z^2 \left(a + 4 + \frac{4}{a}\right).$$ 16. **Conclusion:** Since $$xy = z^2 \left(a + 4 + \frac{4}{a}\right) = z(y + 2x),$$ we have shown that $$xy = z(y + 2x).$$