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If \(\log _7 2=\lambda\), then the value of \(\log _{49}(28)\) is
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The correct answer is:
\(\frac{1}{2}(2 \lambda+1)\)
Hints: \(\log _{49} 28=\log _{72} 4 \times 7\)
\(=\frac{1}{2}\left[2 \log _7 2+\log _7 7\right]=\frac{1}{2}[2 \lambda+1]\)
\(=\frac{1}{2}\left[2 \log _7 2+\log _7 7\right]=\frac{1}{2}[2 \lambda+1]\)
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