3 years ago

Condense 2log4 + log3 - log2 + log5

1 answer:

3 0

$2\log4 + \log3 -\log2 + \log5= \\ \\ = \log 2^4+ \log3 -\log2 + \log5 = \\ \\ = \log 16+ \log3 -\log2 + \log5 = \\ \\ = \log\Big(16\cdot 3:2\cdot 5\Big) = \log\Big(\dfrac{16\cdot 3\cdot 5}{2}\Big) = \log(8\cdot 3\cdot 5) = \\ \\ =\log120$

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