Question

How do you solve for x in `log_6 3x-log_6 (x+1)=log_6 1`?

Answer 1

You can use the property of the log:
`log_aM-log_aN=log_a(M/N)`
So you have:
`log_6((3x)/(x+1))=log_6(1)`
`log_6((3x)/(x+1))=0`
Because from the definition od logarithm:
`log_ab=x => a^x=b`
and: `log_6(1)=0`

as for:
`log_6((3x)/(x+1))=0`
again you get:
`((3x)/(x+1))=6^0=1`
`3x=x+1`
`x=1/2`