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

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Aug 21, 2016

Answer:

#x=33#

Explanation:

By definition if #loga=b#, we have #10^b=a#, hence

#log(3x+1)=2#

#hArr(3x+1)=10^2# or

#3x+1=100# or

#3x=100-1# or

#3x=99# or

#x=99/3=33#

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