Question

How do you condense `2[ln 8- ln(x^2+1)]`?

Answer 1

I got as far as: `ln(8/(x^2+1))^2`

Explanation:

We can use two properties of logs to write:
`2[ln(8/(x^2+1))]=`
(change a subtraction of logs into a division of arguments)

and again:
`=ln(8/(x^2+1))^2`
(transport a number multiplying the log as exponent of the argment).