TS EAMCET · Maths · Basic of Mathematics
\(\log _4 2-\log _8 2+\log _{16} 2-\ldots\) is equal to
- A \(e^2\)
- B \(\log _e 2\)
- C \(1+\log _e 3\)
- D \(1-\log _e 2\)
Answer & Solution
Correct Answer
(D) \(1-\log _e 2\)
Step-by-step Solution
Detailed explanation
\(\log _4 2-\log _8 2+\log _{16} 2-\ldots\) \(=\frac{1}{\log _2 4}-\frac{1}{\log _2 8}+\frac{1}{\log _2 16}-\ldots\) \(\left\{\because \log _b a=\frac{1}{\log _a b}\right\}\) \(=\frac{1}{\log _2(2)^2}-\frac{1}{\log _2(2)^3}+\frac{1}{\log _2(2)^4}-\ldots\)…
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