(1) Find the value of \( \log 2512 \) ? (2) Expand \( \log \frac{x 2 y 3}{z^{2}} \) (3) If \( 5^{x+1}=7^{1-x} \) Find \( X \) value. (4) If \( x^{2}+y^{2}=27 x y \), Show that \( \log \left(\frac{x-y}{5}\right)=\frac{1}{2}(\log x+\log y) \)

To find the value of \( \log 2512 \), you can use its prime factorization. By breaking it down, \( 2512 = 2^4 \times 157 \), thus \( \log 2512 = \log(2^4 \times 157) = 4 \log 2 + \log 157 \). When expanding \( \log \frac{x^2 y^3}{z^2} \), use the properties of logarithms. You would get \( \log x^2 + \log y^3 - \log z^2 = 2 \log x + 3 \log y - 2 \log z \). This helps simplify the expression nicely! For solving \( 5^{x+1} = 7^{1-x} \), take logs on both sides. After manipulation, this gives you an equation in terms of \( x \). Solving yields \( x = \frac{\log 7 - 1}{\log 5 + \log 7} \). In the equation \( x^2 + y^2 = 27xy \), rearranging leads to \( \frac{x-y}{5} \) appearing naturally. By manipulating the terms and applying logarithmic identities, you can derive \( \log \left(\frac{x-y}{5}\right)=\frac{1}{2}(\log x+\log y) \); it’s like piecing together a mathematical puzzle!