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If $\log_{10}y + 3\log_{10}x \geq \log_{10}x$ , express y i...

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.

$y \geq \frac{1}{x^{2}}$

View answer & explanation

Correct answer is D

$\log_{10}y + 3\log_{10}x \geq \log_{10}x$

$\implies \log_{10}y \geq \log_{10}x - 3 \log_{10}x$

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

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

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