Math Problem Statement

Simplify the expression \( \frac{8^x \cdot 6^{2x-1}}{\frac{x}{3} \cdot 16^x \cdot 27} \).

Solution

The given expression appears to be:

8x62x1x316x27.\frac{8^x \cdot 6^{2x-1}}{\frac{x}{3} \cdot 16^x \cdot 27}.

Let's simplify it step by step:


Step 1: Simplify the numerator and denominator

Numerator: 8x62x18^x \cdot 6^{2x-1}
Denominator: x316x27\frac{x}{3} \cdot 16^x \cdot 27.

We will break this into smaller parts.


Step 2: Rewrite terms with the same bases (if possible)

  • For 8x8^x and 16x16^x:
    8=238 = 2^3 and 16=2416 = 2^4, so: 8x=(23)x=23x,16x=(24)x=24x.8^x = (2^3)^x = 2^{3x}, \quad 16^x = (2^4)^x = 2^{4x}.

  • For 62x16^{2x-1}:
    No simplification needed.

  • For 2727:
    27=3327 = 3^3.


Step 3: Combine terms

The expression becomes: 23x62x1x324x33.\frac{2^{3x} \cdot 6^{2x-1}}{\frac{x}{3} \cdot 2^{4x} \cdot 3^3}.


Step 4: Simplify the fraction

  1. Combine the coefficients and simplify powers of 22 and 33: Numerator: 23x62x1.\text{Numerator: } 2^{3x} \cdot 6^{2x-1}. Denominator: x324x33.\text{Denominator: } \frac{x}{3} \cdot 2^{4x} \cdot 3^3.

  2. Simplify the x3\frac{x}{3} factor: Multiply everything in the denominator by 33 to eliminate the fraction.

    This gives: 323x62x1x24x34.\frac{3 \cdot 2^{3x} \cdot 6^{2x-1}}{x \cdot 2^{4x} \cdot 3^4}.

  3. Separate the terms:

    • 23x2^{3x} and 24x2^{4x}:
      23x24x=2x.\frac{2^{3x}}{2^{4x}} = 2^{-x}.

    • 33 and 343^4:
      334=33.\frac{3}{3^4} = 3^{-3}.

    • Combine the 62x16^{2x-1}:
      62x16^{2x-1} remains as it is.

    The simplified result is: 62x12x33x.\frac{6^{2x-1} \cdot 2^{-x} \cdot 3^{-3}}{x}.


Final Simplification

Substitute 6=236 = 2 \cdot 3 into 62x16^{2x-1}: 62x1=(23)2x1=22x132x1.6^{2x-1} = (2 \cdot 3)^{2x-1} = 2^{2x-1} \cdot 3^{2x-1}.

Now the expression becomes: 22x132x12x33x.\frac{2^{2x-1} \cdot 3^{2x-1} \cdot 2^{-x} \cdot 3^{-3}}{x}.

Simplify the powers:

  • For 22x12x2^{2x-1} \cdot 2^{-x}:
    22x12x=2x1.2^{2x-1} \cdot 2^{-x} = 2^{x-1}.

  • For 32x1333^{2x-1} \cdot 3^{-3}:
    32x133=32x4.3^{2x-1} \cdot 3^{-3} = 3^{2x-4}.

Thus, the final expression is: 2x132x4x.\frac{2^{x-1} \cdot 3^{2x-4}}{x}.


Final Answer:

2x132x4x.\frac{2^{x-1} \cdot 3^{2x-4}}{x}.

Would you like further clarification or help with specific steps? 😊


Related Questions:

  1. How do we handle fractional exponents in simplifications?
  2. What happens if x=0x = 0 in this expression?
  3. How can 62x16^{2x-1} be further simplified using logarithms?
  4. Why do powers of 2 and 3 simplify independently here?
  5. Can we evaluate this expression for specific values of xx?

Tip:

Always break down terms into their prime bases (like 22 and 33) to simplify exponential expressions effectively.

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Math Problem Analysis

Mathematical Concepts

Exponents
Fraction simplification
Prime factorization

Formulas

a^m \cdot a^n = a^{m+n}
a^m / a^n = a^{m-n}
(a^m)^n = a^{m \cdot n}

Theorems

Laws of exponents
Prime factorization for base simplification

Suitable Grade Level

Grades 10-12