If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.

  • A \(y \geq \frac{1}{x}\)
  • B \(y \leq \frac{1}{x}\)
  • C \(y \leq \frac{1}{x^{2}}\)
  • D \(y \geq \frac{1}{x^{2}}\)
waec 2016furthermathematics

The correct answer is D. \(y \geq \frac{1}{x^{2}}\)

\(\log_{10}y + 3\log_{10}x \geq \log_{10}x\)

\(\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x \)

\(\log_{10}y \geq -2\log_{10}x = \log_{10}y \geq \log_{10}x^{-2}\)

\(\log_{10}y \geq \log_{10}(\frac{1}{x^{2}}) \implies y \geq \frac{1}{x^{2}}\)

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