Question

How do you solve `ln(x+6)+ln(x-6)=0`?

Answer 1

I found: `x=sqrt(37)=6.082`

Explanation:

You can start by using the rule of logs:
`loga+logb=log(a*b)`

In your case you get:
`ln[(x+6)*(x-6)]=0`

Now you can use the definition of log as:
`log_ax=b -> x=a^b` remembering that the natural log is: `lnx=log_ex`

so:
`(x+6)(x-6)=e^0`
`(x+6)(x-6)=1`
rearranging:
`x^2-36=1`
`x^2=37`
`x=+-sqrt(37)=+-6.082`
The negative `x` cannot be accepted (substituted back it gives a negative argument in the original logs).