\( p ^ { i l - ( k - 1 ) } \cdot ( \frac { - 3 } { p } ) ^ { k - 1 } = p ^ { 5 } \)

To solve the equation \[ p^{il - (k - 1)} \cdot \left( \frac{-3}{p} \right)^{k - 1} = p^{5}, \] let's break it down step by step. 1. **Rewrite the left side**: The term \(\left( \frac{-3}{p} \right)^{k - 1}\) can be rewritten as \((-3)^{k - 1} \cdot p^{-(k - 1)}\). Therefore, the equation becomes: \[ p^{il - (k - 1)} \cdot (-3)^{k - 1} \cdot p^{-(k - 1)} = p^{5}. \] 2. **Combine the powers of \(p\)**: The left side can be simplified by combining the powers of \(p\): \[ p^{il - (k - 1) - (k - 1)} \cdot (-3)^{k - 1} = p^{5}. \] This simplifies to: \[ p^{il - 2(k - 1)} \cdot (-3)^{k - 1} = p^{5}. \] 3. **Equate the powers of \(p\)**: Since the bases are the same, we can equate the exponents: \[ il - 2(k - 1) = 5. \] 4. **Solve for \(p\)**: The equation now involves \(il\) and \(k\). We can express \(p\) in terms of \(k\) and \(il\): \[ il - 2k + 2 = 5 \implies il - 2k = 3. \] This gives us a relationship between \(il\) and \(k\). 5. **Consider the term \((-3)^{k - 1}\)**: Since the left side must equal \(p^5\), we also need to consider the term \((-3)^{k - 1}\). For the equation to hold, \((-3)^{k - 1}\) must equal 1, which occurs when \(k - 1\) is even. Thus, \(k\) must be odd. In summary, we have: - The relationship \(il - 2k = 3\). - \(k\) must be odd for the equation to hold. If you have specific values for \(i\) and \(l\), we can further solve for \(k\) and \(p\).