How do you solve this maths question? #6x+3log2=2x-1log9#

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#6x+3log2=2x-1log9#

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Answer 1 · 2017-10-01T17:40:19

#6x + 3log2 = 2x - log9# can be written as:

#6x + log(2^3) = 2x - log9#

i.e. #6x + log8 = 2x - log9#

Hence, #6x - 2x = -log8 - log9#

or, #6x - 2x = -(log8 + log9)#

so, #4x = -log72#

i.e. #x = -log(72)/4#

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