If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
Question 1 Report
If \(\log_{10}y + 3\log_{10}x \geq \log_{10}x\), express y in terms of x.
Answer Details
We can start by using the logarithmic rule that states: \(\log_{a}b + \log_{a}c = \log_{a}(bc)\).
So, we can rewrite the given inequality as follows:
\[\log_{10}y + 3\log_{10}x \geq \log_{10}x\]
\[\Rightarrow \log_{10}(yx^{3}) \geq \log_{10}x\]
\[\Rightarrow yx^{3} \geq x\]
Dividing both sides by \(x\) (since \(x > 0\)), we get:
\[yx^{2} \geq 1\]
\[\Rightarrow y \geq \frac{1}{x^{2}}\]
Therefore, the answer is \(y \geq \frac{1}{x^{2}}\).
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