Find \( P \) in. \( p^{11-(k-1)} \cdot\left(\frac{-3}{p}\right)^{k-1}=p^{5} \)
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Bonus Knowledge
To find \( P \) in the equation \( p^{11-(k-1)} \cdot\left(\frac{-3}{p}\right)^{k-1}=p^{5} \), we can start by simplifying the left side. First, rewrite \( \left(\frac{-3}{p}\right)^{k-1} \): \[ \left(\frac{-3}{p}\right)^{k-1} = (-3)^{k-1} \cdot p^{-(k-1)} \] Now substitute this back into the equation: \[ p^{11-(k-1)} \cdot (-3)^{k-1} \cdot p^{-(k-1)} = p^{5} \] Combine the powers of \( p \): \[ (-3)^{k-1} \cdot p^{11 - (k-1) - (k-1)} = p^{5} \] This simplifies to: \[ (-3)^{k-1} \cdot p^{11 - 2(k-1)} = p^{5} \] Thus, you can write: \[ (-3)^{k-1} \cdot p^{11 - 2k + 2} = p^{5} \] Next, equate the powers of \( p \): \[ 11 - 2k + 2 = 5 \] Combine like terms: \[ 13 - 2k = 5 \] Now solve for \( k \): \[ 13 - 5 = 2k \implies 8 = 2k \implies k = 4 \] Substituting \( k = 4 \) back into the equation: \[ (-3)^{4-1} \cdot p^{11 - 2 \cdot 4 + 2} = p^{5} \] This gives us: \[ (-3)^{3} \cdot p^{11 - 8 + 2} = p^{5} \] or: \[ -27 \cdot p^{5} = p^{5} \] Now divide both sides by \( p^{5} \) (assuming \( p \neq 0 \)): \[ -27 = 1 \] This is clearly a contradiction, indicating that the manipulation may have led us to assume that there are conditions under which \( k \) and \( p \) can exist. Thus, the equation does not hold for any non-zero \( p \) under real conditions or needs to be reconsidered if you have specific constraints or alternate contexts. In summary, there is no value for \( P \) that satisfies the given equation for any real number \( p \) since we reached a contradiction. \( P \) can only exist if additional constraints are defined.