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