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

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Answer 1 · 2016-01-13T03:20:29

#x=33#

This can be rewritten as

#log_10(3x+1)=2#

To undo the logarithm with base #10#, exponentiate both sides with a base of #10#.

#color(red)(cancel(color(black)(10^(log_10))))""^((3x+1))=10^2#

#3x+1=100#

#3x=99#

#x=33#

Answer 2 · 2016-01-13T16:57:23

Just another way of writing exactly the same thing!

#x=99/3=33#

Given: #log_10(3x+1)=2#

2 is the index that 10 has to be raised to give 3x+1. So we have:

#10^2=3x+1#

#100 =3x+1#

#3x=99#

#x=99/3=33#