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

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:

\[
3 \cdot 2^x - 4 \cdot 2^{y-2} \div 2^x - \frac{2^{x-1}}{16}
\]

### Step 2: Simplify the division terms
\[
\frac{2^{y-2}}{2^x} = 2^{(y-2) - x} = 2^{y-x-2}
\]
\[
\frac{2^{x-1}}{16} = \frac{2^{x-1}}{2^4} = 2^{(x-1) - 4} = 2^{x-5}
\]

### Step 3: Substitute these results back into the expression
\[
3 \cdot 2^x - 4 \cdot 2^{y-x-2} - 2^{x-5}
\]

### Final Simplified Expression:
\[
3 \cdot 2^x - 4 \cdot 2^{y-x-2} - 2^{x-5}
\]

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!

This page provides a detailed step-by-step solution to the expression root(3)(27⋅)2^(^(x))- root(3)(64⋅)2^(y-2)÷2^(x)-2^(x-1)÷16, focusing on simplifying cube roots and using exponent rules. Suitable for high school students in Grades 10-12.

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