Question

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

Answer 1

I found: `b=+-4`

Explanation:

You can start using a property of logs:
`logx+logy=logxy`
so you get:
`log_6[2(b^2+2)]=2`
using the definition of log:
`log_ax=b -> x=a^b` you get:
`2(b^2+2)=6^2=36`
`b^2+2=36/2=18`
`b^2=18-2=16`
so that `b=+-sqrt(16)=+-4`