How do I find the logarithm #log_(2/3)(8/27)#?

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How do I find the logarithm #log_(2/3)(8/27)#?

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Answer 1 · 2018-03-05T09:24:11

#color(red)3#

Explanation:

(1)#log_aX^n=nlog_aX#
(2)#log_aa=1#
Now,
#log_(2/3)(8/27)=log_(2/3)(2^3/3^3)=log_(2/3)(2/3)^color(red)(3)# [Applying(1), for n=3]
#=(3)*log_color(red)((2/3))(2/3)=3(1)=3#, [Applying(2) for # a=2/3# ]

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How do I find the logarithm #log_(2/3)(8/27)#?

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