If log72 = λ, then the value of log49(28) is(a) \(\frac{1}{2}\) (2λ

1.	If log72 = λ, then the value of log49(28) is(a) \(\frac{1}{2}\) (2λ + 1)(b) (2λ + 1) (c) 2 (2λ + 1) (d) \(\frac{3}{2}\) (2λ + 1)
Answer» (a) \(\frac{1}{2}\) (2λ + 1) log₄₉(28) = log_7² (7 x 2²) = log_7² 7 + log_7² 2² = \(\frac{1}{2}\) log₇ 7 + \(\frac{2}{2}\) log₇ 2 \(\bigg[\)Using log_aⁿ\(x\) = \(\frac{1}{n}\) log_a x, log_aⁿ (x^m) = \(\frac{m}{n}\) log_a x\(\bigg]\) = \(\frac{1}{2}\) + λ = \(\frac{1}{2}\)(2λ + 1).

If log72 = λ, then the value of log49(28) is(a) \(\frac{1}{2}\) (2λ + 1)(b) (2λ + 1) (c) 2 (2λ + 1) (d) \(\frac{3}{2}\) (2λ + 1)

Answer»

(a) \(\frac{1}{2}\) (2λ + 1)

log₄₉(28) = log_7² (7 x 2²) = log_7² 7 + log_7² 2²

= \(\frac{1}{2}\) log₇ 7 + \(\frac{2}{2}\) log₇ 2

\(\bigg[\)Using log_aⁿ\(x\) = \(\frac{1}{n}\) log_a x, log_aⁿ (x^m) = \(\frac{m}{n}\) log_a x\(\bigg]\)

= \(\frac{1}{2}\) + λ = \(\frac{1}{2}\)(2λ + 1).