How do you solve #Log 2 + log(4x – 1) = 3#?

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Answer 1 · 2018-03-18T09:13:53

#log2 +log(4x-1)=3#

Or, #log(2×(4x-1))=3log10=log 10^3#

Or, #8x-2=1000#

So, #8x=1002#

Or, #x=125.25#

Answer 2 · 2018-03-18T09:17:08

#x=125.25#

the logs can be combined using the rule:
#loga+logb=logab#
so

#log(2(4x-1))=3#
#log(8x-2)=3#

to get rid of the log, we put both sides to the power of ten
#10^(log(8x-2))=10^3#

the ten and the log cancel out, leaving
#8x-2=1000#
#8x=1002#
#x=125.25#