Basic Maths and Logarithm
Properties of Logarithms
JEE Advanced
Grade 11
Question:
If $\log_7 2 = \lambda$, then $\log_{49} 28$ equals:
(1) $\frac{1 + 2\lambda}{4}$
(2) $\frac{1 + 2\lambda}{2}$
(3) $\frac{1 + 2\lambda}{3}$
(4) $\frac{2 + \lambda}{4}$
Step-by-Step Solution
Key Concept: $\log_{49} 28 = \frac{1}{2} \log_7 28 = \frac{1}{2} \log_7 (4 \times 7) = \frac{1}{2} (\log_7 4 + 1) = \frac{1}{2} (2 \log_7 2 + 1) = \frac{1+2\lambda}{2}$.
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Correct Answer: (2)