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Find \( P \) in. \( p^{11-(k-1)} \cdot\left(\frac{-3}{p}\right)^{k-1}=p^{5} \)

Ask by Lowe Page. in South Africa
Feb 19,2025

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Tutor-Verified Answer

Answer

The solutions for \( p \) are \( p = 0 \) or \( p = (-3)^{-\frac{1}{6}} \).

Solution

Solve the equation by following steps: - step0: Solve for \(p\): \(p^{11-\left(k-1\right)}\left(\frac{-3}{p}\right)^{k-1}=p^{5}\) - step1: Simplify: \(p^{12-k}\left(-\frac{3}{p}\right)^{k-1}=p^{5}\) - step2: Expand the expression: \(\left(-3\right)^{k-1}p^{13-2k}=p^{5}\) - step3: Rewrite the expression: \(p^{13-2k}=\left(-3\right)^{-k+1}p^{5}\) - step4: Move the expression to the left side: \(p^{13-2k}-\left(-3\right)^{-k+1}p^{5}=0\) - step5: Factor the expression: \(p^{5}\left(p^{8-2k}+3^{-k+1}\right)=0\) - step6: Separate into possible cases: \(\begin{align}&p^{5}=0\\&p^{8-2k}+3^{-k+1}=0\end{align}\) - step7: Set the base equal to 0: \(\begin{align}&p=0\\&p=\left(-3^{-k+1}\right)^{\frac{1}{8-2k}}\end{align}\) To solve the equation \[ p^{11-(k-1)} \cdot \left(\frac{-3}{p}\right)^{k-1} = p^{5}, \] we can simplify and analyze it step by step. 1. **Rewrite the equation**: The left side can be rewritten as: \[ p^{11 - (k - 1)} \cdot \left(-3\right)^{k - 1} \cdot p^{-(k - 1)} = p^{11 - k + 1} \cdot (-3)^{k - 1}. \] Thus, the equation becomes: \[ p^{12 - k} \cdot (-3)^{k - 1} = p^{5}. \] 2. **Equate the powers of \( p \)**: We can separate the terms involving \( p \): \[ p^{12 - k} = p^{5} \cdot \frac{1}{(-3)^{k - 1}}. \] 3. **Set the powers equal**: This gives us two cases to consider: - Case 1: \( p = 0 \) - Case 2: \( 12 - k = 5 \) 4. **Solve for \( k \)**: From \( 12 - k = 5 \): \[ k = 7. \] 5. **Substituting \( k \) back**: Now substituting \( k = 7 \) back into the equation gives: \[ p^{12 - 7} = p^{5} \cdot \frac{1}{(-3)^{6}}. \] Simplifying this, we have: \[ p^{5} = p^{5} \cdot \frac{1}{729}. \] 6. **Final solutions**: Thus, the solutions for \( p \) are: \[ p = 0 \quad \text{or} \quad p = (-3)^{\frac{1}{8 - 2k}}. \] Substituting \( k = 7 \) into the second solution gives: \[ p = (-3)^{\frac{1}{8 - 14}} = (-3)^{-\frac{1}{6}}. \] In conclusion, the solutions for \( p \) are: \[ p = 0 \quad \text{or} \quad p = (-3)^{-\frac{1}{6}}. \]

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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.

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