Evaluate log x³+log5x=5log2-log (2/5)?

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Answer 1 · 2018-01-11T08:08:53

#x = 2#

Before solving this type of question, you must first remember some logarithmic properties,

#logx^y = ylogx#
#logx*y = logx + logy#
#log(x/y) = logx - logy#

#logx^3 + log5x = 5log2 - log(2/5)#

By applying appropriate properties,

#=> 3logx + log5 + logx = 5log2 - (log2 - log5)#
#=> 4logx = 4log2#

Hence,
#x = 2#

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