6. (x+1) log _(15) y=(2 y)/(5), log _(25) x=(5)/(2 y) ; p=y^2-2 x^2

Let's break down this problem step-by-step.<br /><br />**Understanding the Problem**<br /><br />We're given two equations involving logarithms and asked to find the value of an expression 'P' in terms of 'x' and 'y'. Here's what we need to keep in mind:<br /><br />* **Logarithm Base 0:** The logarithm with base 0 is undefined. This means the first equation is not valid. We'll need to assume there's a typo and work with a different base.<br />* **Logarithm Properties:** We'll use the following properties of logarithms:<br /> * `log_a (b) = c` is equivalent to `a^c = b`<br /> * `log_a (b^n) = n * log_a (b)`<br /> * `log_a (b) = log_c (b) / log_c (a)` (Change of Base Formula)<br /><br />**Solving the Equations**<br /><br />Let's assume the first equation should be `(x+1)log_a y = 2y/5` (where 'a' is a valid base).<br /><br />1. **Rewrite the Equations in Exponential Form:**<br /> * `(x+1)log_a y = 2y/5` => `a^(2y/5) = y^(x+1)`<br /> * `log_{25} x = 5/(2y)` => `25^(5/(2y)) = x`<br /><br />2. **Simplify the Second Equation:**<br /> * `25^(5/(2y)) = x` => `(5^2)^(5/(2y)) = x` => `5^(5/y) = x`<br /><br />3. **Express 'x' in terms of 'y':**<br /> * From the simplified second equation, we have `x = 5^(5/y)`<br /><br />4. **Substitute 'x' into the First Equation:**<br /> * `a^(2y/5) = y^(x+1)` => `a^(2y/5) = y^(5^(5/y) + 1)`<br /><br />5. **Solve for 'y':**<br /> * Unfortunately, we can't directly solve for 'y' in this equation. We need more information about the base 'a' to proceed.<br /><br />**Finding the Value of P**<br /><br />Since we can't solve for 'y' explicitly, we can't directly find the value of `P = y^2 - 2x^2`. We need more information about the base 'a' or additional constraints on 'x' and 'y'.<br /><br />**Conclusion**<br /><br />The problem as presented is incomplete. We need to know the correct base for the first logarithm or have additional information to solve for 'y' and find the value of 'P'. <br />