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:

$10^2=3x+1$

$100 =3x+1$

$3x=99$

$x=99/3=33$

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