JEE Advanced · Mathematics · 1. Basic Math
Let \(\left(x_0, y_0\right)\) be the solution of the following equations \((2 x)^{\log 2}=(3 y)^{\log 3}\), \(3^{\log x}=2^{\log y}\), then \(x_0\) is equal to
- A \(\frac{1}{6}\)
- B \(\frac{1}{2}\)
- C \(\frac{1}{2}\)
- D 6
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Taking log on both sides,
\(\log 2 \cdot \log (2 x)=\log 3(\log 3 y)\)
\(\Rightarrow \log 2\{\log 2+\log x\} \)
\(=\log 3\{\log 3+\log y\} \)
\(\text {and } \log x \cdot \log 3=\log y \log 2 \)
\(\log y=\frac{\log x \cdot \log 3}{\log 2}\)
From Eqs. (i) and (ii), we get
\(\log 2\{\log 2+\log x\} \)
\(=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \)
\(\Rightarrow (\log 2)^2+\log 2 \cdot \log x \)
\(=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \)
\(\Rightarrow \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \)
\(=(\log 2)^2-(\log 3)^2\)
\(\Rightarrow \log x=-\log 2=\log 2^{-1}\)
\(\therefore x=\frac{1}{2}\)
\(\log 2 \cdot \log (2 x)=\log 3(\log 3 y)\)
\(\Rightarrow \log 2\{\log 2+\log x\} \)
\(=\log 3\{\log 3+\log y\} \)
\(\text {and } \log x \cdot \log 3=\log y \log 2 \)
\(\log y=\frac{\log x \cdot \log 3}{\log 2}\)
From Eqs. (i) and (ii), we get
\(\log 2\{\log 2+\log x\} \)
\(=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \)
\(\Rightarrow (\log 2)^2+\log 2 \cdot \log x \)
\(=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \)
\(\Rightarrow \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \)
\(=(\log 2)^2-(\log 3)^2\)
\(\Rightarrow \log x=-\log 2=\log 2^{-1}\)
\(\therefore x=\frac{1}{2}\)
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