Question

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 1

(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).