Question

How do you solve `log(x+1) + log(x-1) = 1`?

Answer 1

I found: `x=+sqrt(11)`

Explanation:

You can use the rule of logs that relates the sum of logs of same base and the multiplication of the integrand as:
`log_ax+log_ay=log_a(x*y)`
and get:

`log_a[(x+1)(x-1)]=1` using the definition of log:

`(x+1)(x-1)=a^1` (1)

Now it depends upon the base `a` of your logs....and also if it the same for both the original ones!
If `a=10` then you have:
`x^2-1=10`
`x^2=11`
`x=+-sqrt(11)`
Where I accept the positive only, `x=+sqrt(11)`

If `a` isn't `10` simply insert the right value into (1) and solve.