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Explain two non-verbal cues you would use to make the performance of the poem effective

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Answer

117\mathbf{\frac{11}{7}}

Step-by-step solution

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Here's the continuation of the previous answer:

FURTHER LOGARITHMS

3. Find the value of xx given that log⁡(15−5x)−1=log⁡(3x−4)\log(15 - 5x) - 1 = \log(3x - 4). (Assuming base 10 for log⁡\log)

  1. Express 1 as a logarithm with base 10.

    1=log⁡10101 = \log_{10} 10
  2. Rewrite the equation.

    log⁡(15−5x)−log⁡10=log⁡(3x−4)\log(15 - 5x) - \log 10 = \log(3x - 4)
  3. Apply the logarithm property log⁡a−log⁡b=log⁡(ab)\log a - \log b = \log \left(\frac{a}{b}\right) on the left side.

    log⁡(15−5x10)=log⁡(3x−4)\log \left(\frac{15 - 5x}{10}\right) = \log(3x - 4)
  4. Equate the arguments of the logarithms.

    15−5x10=3x−4\frac{15 - 5x}{10} = 3x - 4
  5. Solve the resulting algebraic equation.

    15−5x=10(3x−4)15 - 5x = 10(3x - 4) 15−5x=30x−4015 - 5x = 30x - 40 15+40=30x+5x15 + 40 = 30x + 5x 55=35x55 = 35x x=5535x = \frac{55}{35} x=117x = \frac{11}{7}
  6. Check for valid solutions.

    The arguments of logarithms must be positive.
    For log⁡(15−5x)\log(15 - 5x): 15−5x>0  ⟹  15>5x  ⟹  x<315 - 5x > 0 \implies 15 > 5x \implies x < 3.
    For log⁡(3x−4)\log(3x - 4): 3x−4>0  ⟹  3x>4  ⟹  x>433x - 4 > 0 \implies 3x > 4 \implies x > \frac{4}{3}.
    Our solution x=117≈1.57x = \frac{11}{7} \approx 1.57. This value satisfies both x<3x < 3 and x>43x > \frac{4}{3}.

    x=117x = \boxed{\mathbf{\frac{11}{7}}}
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