Math Problem Statement

are you sure? i tested numbers as inputs and they always gave the same result for both sides.

Solution

Let's re-examine the expression more carefully, considering the potential mistake in the previous explanation. We'll analyze both sides of the equation to see if they indeed match for any given numbers.

Given Equation:

2x+(2x1)3(2y1)3=2x+y132^x + \frac{(2^x - 1)}{3} \cdot \frac{(2^y - 1)}{3} = \frac{2^{x+y} - 1}{3}

Step-by-Step Analysis:

  1. Left-hand side (LHS):

2x+(2x1)3(2y1)32^x + \frac{(2^x - 1)}{3} \cdot \frac{(2^y - 1)}{3}

We can rewrite the product as: (2x1)(2y1)9\frac{(2^x - 1)(2^y - 1)}{9}

So the LHS becomes: 2x+(2x1)(2y1)92^x + \frac{(2^x - 1)(2^y - 1)}{9}

Expand (2x1)(2y1)(2^x - 1)(2^y - 1): (2x1)(2y1)=2x2y2x2y+1=2x+y2x2y+1(2^x - 1)(2^y - 1) = 2^x \cdot 2^y - 2^x - 2^y + 1 = 2^{x+y} - 2^x - 2^y + 1

Substituting this back into the LHS: 2x+2x+y2x2y+192^x + \frac{2^{x+y} - 2^x - 2^y + 1}{9}

  1. Right-hand side (RHS): 2x+y13\frac{2^{x+y} - 1}{3}

  2. Comparing the simplified forms:

Checking LHS:

2x+2x+y2x2y+192^x + \frac{2^{x+y} - 2^x - 2^y + 1}{9}

Multiply everything by 9 to get a common denominator: 92x+2x+y2x2y+19\frac{9 \cdot 2^x + 2^{x+y} - 2^x - 2^y + 1}{9}

Simplify the numerator: 82x+2x+y2y+19\frac{8 \cdot 2^x + 2^{x+y} - 2^y + 1}{9}

Checking RHS:

Multiply RHS by 3: 3(2x+y1)3=2x+y1\frac{3 \cdot (2^{x+y} - 1)}{3} = 2^{x+y} - 1

Now, considering the feedback, let's check with numbers directly.

Let's test with x=1x = 1 and y=2y = 2:

  • 21+(211)3(221)32^1 + \frac{(2^1 - 1)}{3} \cdot \frac{(2^2 - 1)}{3} 2+(21)3(41)3=2+1333=2+13=2+19=2.11112 + \frac{(2 - 1)}{3} \cdot \frac{(4 - 1)}{3} = 2 + \frac{1}{3} \cdot \frac{3}{3} = 2 + \frac{1}{3} = 2 + \frac{1}{9} = 2.1111
  • 21+213\frac{2^{1+2} - 1}{3} 2313=813=73=2.3333\frac{2^3 - 1}{3} = \frac{8 - 1}{3} = \frac{7}{3} = 2.3333

When substituting with the numbers, the left and right-hand sides do not match.

Let's test with another pair x=2x = 2 and y=1y = 1:

  • 22+(221)3(211)32^2 + \frac{(2^2 - 1)}{3} \cdot \frac{(2^1 - 1)}{3} 4+(41)3(21)3=4+3313=4+1=4.11114 + \frac{(4 - 1)}{3} \cdot \frac{(2 - 1)}{3} = 4 + \frac{3}{3} \cdot \frac{1}{3} = 4 + 1 = 4.1111

  • 22+113\frac{2^{2+1} - 1}{3} 2313=813=73=2.3333\frac{2^3 - 1}{3} = \frac{8 - 1}{3} = \frac{7}{3} = 2.3333

Again, we see different results. This confirms that the expressions are not equivalent.

Would you like any more details on this solution or have any questions? Here are some related questions you might be interested in:

  1. How to simplify complex fractions involving exponential terms?
  2. What are the common properties of exponents?
  3. How to factor expressions involving exponents?
  4. What are the steps for solving exponential equations?
  5. How to verify if two algebraic expressions are equivalent?

Tip: When dealing with exponentiation and fractional expressions, always try to simplify step by step and check if you can factor out common terms.

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Math Problem Analysis

Mathematical Concepts

Exponentiation
Fractional Expressions
Equations

Formulas

Exponential expressions
Fractional equations

Theorems

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Suitable Grade Level

Advanced High School