Question

Prove that any base log (31/16) – log (8/9) + log (128/243) = log (3)​

Answer 1

Step-by-step explanation:

I think first term is log 21/16 instead of log 31/16

log a - log b + log c = log a÷b×c

log 21/16 ÷ 8/9 ×128/243

= log 21/16 × 9/8 × 128/243

=log 21×9×128/16×8×243

= log 21×3×(3×128)/(128×3)×21

= log 21×3/21 = log 3