bantu aku,bagaimana cara penyelesaian logaritma ini

LogAritMa

(a^n)'log (b^m) = (m/n) . a'log b

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(16'log 2 + 3'log 4 . √2 'log 9) / (2'log 3 - 2'log 12)
= ((2⁴)'log 2 + 3'log 2² . (2^1/2)'log 3²) / (2'log (3/12))
= (1/4 + (2/(1/2))) / (2'log 2^-2)
= (1/4 + 4) / (-2)
= -17/8