If ₁0}y + 3₁0}x ≥ ₁0}x, express y in terms of x.

Correct Answer is : y≥ 1x2y ≥ \;{1}{x²}y≥x21

Correct Answer is : y≥ 1x2y ≥ \;{1}{x²}y≥x21

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WAEC Further Mathematics 2016 Paper

Question : 12

Total: 40

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

, express y in terms of x.

Solution:

Concept:

Use the power rule of logarithms to simplify the inequality, then compare the arguments because

\log_{10}x

is increasing.

Explanation:

Given:

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

Subtract

3\log_{10}x

from both sides:

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

Simplify the right-hand side:

\log_{10}x - 3\log_{10}x = -2\log_{10}x = \log_{10}(x^{-2})

Thus,

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

Since

x^{-2} = \frac{1}{x^{2}}

, we have

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

Because logarithms are increasing, the inequality carries over to the arguments:

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

Answer:

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

This is Option D.

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