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2(3x+1)+7=32x2(3^x+1) +7= 3^{2x}

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Q. 2(3x+1)+7=32x2(3^x+1) +7= 3^{2x}
  1. Simplify Equation: First, let's simplify the equation by distributing the 22 into the parentheses: 2×3x+2×1+7=3(2x)2 \times 3^x + 2 \times 1 + 7 = 3^{(2x)}.\newlineSo we get 2×3x+2+7=3(2x)2 \times 3^x + 2 + 7 = 3^{(2x)}.
  2. Combine Like Terms: Now, combine like terms on the left side: 2×3x+9=32x2 \times 3^x + 9 = 3^{2x}.
  3. Isolate Terms with x: Subtract 99 from both sides to isolate terms with xx on one side: 2×3x=32x92 \times 3^x = 3^{2x} - 9.
  4. Write in Same Base: We can write 32x3^{2x} as (3x)2(3^x)^2 to have the same base: 2×3x=(3x)292 \times 3^x = (3^x)^2 - 9.
  5. Set Equation to Zero: Let's set this equation equal to zero: (3x)223x9=0(3^x)^2 - 2 \cdot 3^x - 9 = 0.
  6. Substitution Method: Now, we can use a substitution method. Let's let u=3xu = 3^x. So our equation becomes u22u9=0u^2 - 2u - 9 = 0.
  7. Factor Quadratic Equation: We can factor this quadratic equation: (u3)(u+3)=0(u - 3)(u + 3) = 0.