Matematika

Pertanyaan

2log96 + 3/4 2log256 - 2/3 2log64 - 2log6

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  • [tex] {}^{2} log96 + \frac{3}{4} {}^{2} log256 - \frac{2}{3} {}^{2} log64 - {}^{2} log6 \\ = ( \frac{3}{4} - \frac{2}{3} ) {}^{2} log( \frac{96 \times 256}{64 \times 6} ) \\ = \frac{1}{12} {}^{2} log64 \\ = \frac{1}{12} {}^{2} log( {2}^{6} ) \\ = \frac{6}{12} {}^{2} log2 \\ = \frac{1}{2} \times 1 \\ = \frac{1}{2} [/tex]

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