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Simplify: \( \begin{array}{l}\text { QUESTION 2 } \\ \frac{2^{x-3}-3 \cdot 2^{x-1}}{2^{x-2}} \\ \\ \frac{2^{2013}-2.3 \cdot 2^{2001}}{2^{x} \cdot 2^{-2}} \\ \frac{2^{2001+2^{2013}-6 \cdot 2^{2011}}-2.3 .2^{200}}{2^{x}-2^{-2}} \\ 2^{2001}\left(2^{2}-2.3\right)\end{array} \)

Ask by Erickson Mcguire. in South Africa
Mar 09,2025

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The simplified result is \( -\frac{11}{2} \).

Solución

We wish to “simplify” \[ \frac{2^{x-3}-3\cdot 2^{x-1}}{2^{x-2}}. \] A natural way is to work only on the numerator and denominator that depend on the same power \(2^{x}\). **Step 1. Rewrite the numerator** In the numerator the two terms are \[ 2^{x-3}\quad\text{and}\quad 3\cdot 2^{x-1}. \] Write both in “base” \(2^{x-3}\). Notice that \[ 2^{x-1} = 2^{x-3}\cdot 2^2. \] Thus the numerator becomes \[ 2^{x-3} - 3\cdot\Bigl(2^{x-3}\cdot 2^2\Bigr) = 2^{x-3}\Bigl(1 - 3\cdot 2^2\Bigr). \] Since \(2^2=4\) we have \[ 1-3\cdot 4 = 1-12 = -11. \] So the numerator is \[ -11\cdot 2^{x-3}. \] **Step 2. Write the denominator** The denominator is \[ 2^{x-2}. \] **Step 3. Cancel common factors** Now, writing the full expression we have \[ \frac{-11\cdot 2^{x-3}}{2^{x-2}}. \] Since the powers have the same base, we can combine: \[ \frac{2^{x-3}}{2^{x-2}} = 2^{(x-3)-(x-2)} = 2^{-1}=\frac12. \] Thus the whole expression becomes \[ -11\cdot \frac12 = -\frac{11}{2}\,. \] The simplified result is \[ -\frac{11}{2}\,. \]

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To simplify the expression, let's analyze it step by step. First, consider the expression: \[ \frac{2^{x-3}-3 \cdot 2^{x-1}}{2^{x-2}}. \] We can simplify the numerator: \[ 2^{x-3} - 3 \cdot 2^{x-1} = 2^{x-3} - 3 \cdot 2^{x-1} = 2^{x-3} - 3 \cdot 2^{x-3} \cdot 2^2 = 2^{x-3} (1 - 12) = 2^{x-3} (-11). \] Therefore, the numerator simplifies to: \[ -11 \cdot 2^{x-3}. \] Now, we can simplify the full fraction: \[ \frac{-11 \cdot 2^{x-3}}{2^{x-2}} = -11 \cdot \frac{2^{x-3}}{2^{x-2}} = -11 \cdot 2^{-1} = -\frac{11}{2}. \] Now, for the second expression: \[ \frac{2^{2013}-2\cdot3 \cdot 2^{2001}}{2^{x} \cdot 2^{-2}}. \] Here, we simplify the denominator: \[ 2^{x} \cdot 2^{-2} = 2^{x-2}. \] Now, simplify the numerator: \[ 2^{2013} - 6 \cdot 2^{2001} = 2^{2001}(2^{12} - 6) = 2^{2001}(4096 - 6) = 2^{2001} \cdot 4090. \] So the entire expression becomes: \[ \frac{2^{2001} \cdot 4090}{2^{x-2}} = 2^{2001 - (x-2)} \cdot 4090 = 2^{2003 - x} \cdot 4090. \] Now consider the third expression: \[ \frac{2^{2001 + 2^{2013} - 6 \cdot 2^{2011}} - 2 \cdot 3 \cdot 2^{200}}{2^{x}-2^{-2}}. \] First simplify the term in the exponent: \[ 2^{2001 + 2^{2013} - 6 \cdot 2^{2011}} = 2^{2001 + (2 \cdot 2^{2011} - 6 \cdot 2^{2011})} = 2^{2001 - 4 \cdot 2^{2011}}. \] The denominator simplifies as: \[ 2^{x} - 2^{-2} = 2^{x} - \frac{1}{4} = \frac{4 \cdot 2^x - 1}{4}. \] The complete expression becomes rather complex, making it difficult to simplify without further information about \(x\). Lastly, the entire original question concludes with: \[ 2^{2001} \left(2^{2} - 2 \cdot 3\right) = 2^{2001} \cdot (4 - 6) = 2^{2001} \cdot (-2) = -2^{2002}. \] So the answers from the parts are: 1. The first fraction simplifies to \(-\frac{11}{2}\). 2. The second fraction simplifies to \(2^{2003-x} \cdot 4090\). 3. The third is a difficult expression depending on \(x\). 4. The conclusion returns a negative multiple of \(2^{2001}\). Thus the expression does not directly equal any of the subsequent expressions without specific values for \(x\).

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