Math Problem Statement

root(3)(27⋅)2^(^(x))- root(3)(64⋅)2^(y-2)÷2^(x)-2^(x-1)÷16

Solution

We have the following expression to simplify:

[ \sqrt[3]{27} \cdot 2^x - \sqrt[3]{64} \cdot 2^{y-2} \div 2^x - \frac{2^{x-1}}{16} ]

Step 1: Simplify the cube roots

[ \sqrt[3]{27} = 3 \quad \text{and} \quad \sqrt[3]{64} = 4 ]

Thus, the expression becomes:

32x42y2÷2x2x1163 \cdot 2^x - 4 \cdot 2^{y-2} \div 2^x - \frac{2^{x-1}}{16}

Step 2: Simplify the division terms

2y22x=2(y2)x=2yx2\frac{2^{y-2}}{2^x} = 2^{(y-2) - x} = 2^{y-x-2} 2x116=2x124=2(x1)4=2x5\frac{2^{x-1}}{16} = \frac{2^{x-1}}{2^4} = 2^{(x-1) - 4} = 2^{x-5}

Step 3: Substitute these results back into the expression

32x42yx22x53 \cdot 2^x - 4 \cdot 2^{y-x-2} - 2^{x-5}

Final Simplified Expression:

32x42yx22x53 \cdot 2^x - 4 \cdot 2^{y-x-2} - 2^{x-5}

Let me know if you need further details or a breakdown of any part!

Related Questions:

  1. How do you apply exponent rules when simplifying expressions?
  2. What is the general method for simplifying cube roots?
  3. How does exponent subtraction work when dividing powers?
  4. Why do we express constants like 16 as powers of 2 during simplification?
  5. Can this expression be factored further under specific conditions?

Tip:

When dividing exponents with the same base, subtract the exponents to simplify quickly!

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Math Problem Analysis

Mathematical Concepts

Algebra
Exponentiation
Roots
Simplification

Formulas

Cube root simplification: \sqrt[3]{a}
Exponent division rule: \frac{a^m}{a^n} = a^{m-n}
Power subtraction: a^m \div a^n = a^{m-n}

Theorems

Exponent Rules
Cube Root Theorem

Suitable Grade Level

Grades 10-12