If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
- \(y \geq \frac{1}{x}\)
- \(y \leq \frac{1}{x}\)
- \(y \leq \frac{1}{x^{2}}\)
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\(y \geq \frac{1}{x^{2}}\)
The correct answer is: D
Explanation
\(\log_{10}y + 3\log_{10}x \geq \log_{10}x\)
\(\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x \)
\(\log_{10}y \geq -2\log_{10}x = \log_{10}y \geq \log_{10}x^{-2}\)
\(\log_{10}y \geq \log_{10}(\frac{1}{x^{2}}) \implies y \geq \frac{1}{x^{2}}\)