Question

How do you solve `log_6(b^2+2) + log_6 2=2`?

Answer 1

`b=+-4`

Explanation:

Start by usig the following property of logs:

`log_ab+log_ac=log_a(b*c)`
so you get:

`log_6[2(b^2+2)]=2`

now use the definition of log:
`log_ax=b ->x=a^b`

in your case:
`2(b^2+2)=6^2`
`b^2+2=18`
`b^2=16`
`b=+-sqrt(16)=+-4`